How do I do this in Algebra?

You can use either quadratic formula or completing the square for both problems.

Question Number 1 :
For this equation x^2 + x = 12 , answer the following questions :
A. Find the roots using Quadratic Formula !
B. Use completing the square to find the root of the equation !

Answer Number 1 :
First, we have to turn equation : x^2 + x = 12 , into a*x^2+b*x+c=0 form.
x^2 + x = 12 , move everything in the right hand side, to the left hand side of the equation
<=> x^2 + x - ( 12 ) = 0 , which is the same with
<=> x^2 + x + ( - 12 ) =0 , now open the bracket and we get
<=> x^2 + x - 12 = 0

The equation x^2 + x - 12 = 0 is already in a*x^2+b*x+c=0 form.
In that form, we can easily derive that the value of a = 1, b = 1, c = -12.

1A. Find the roots using Quadratic Formula !
Use abc formula and you get either
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) or x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
Since a = 1, b = 1 and c = -12,
we just need to subtitute the value of a,b and c in the abc formula.
So we get x1 = (-(1) + sqrt( (1)^2 - 4 * (1)*(-12)))/(2*1) and x2 = (-(1) - sqrt( (1)^2 - 4 * (1)*(-12)))/(2*1)
Which make x1 = ( -1 + sqrt( 1+48))/(2) and x2 = ( -1 - sqrt( 1+48))/(2)
Which is the same as x1 = ( -1 + sqrt( 49))/(2) and x2 = ( -1 - sqrt( 49))/(2)
It imply that x1 = ( -1 + 7 )/(2) and x2 = ( -1 - 7 )/(2)
We get following answers x1 = 3 and x2 = -4

1B. Use completing the square to find the root of the equation !
x^2 + x - 12 = 0 ,divide both side with 1
So we get x^2 + x - 12 = 0 ,
The coefficient of x is 1
We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = 1/2 = 0.5
So we have make the equation into x^2 + x + 0.25 - 12.25 = 0
Which is the same with ( x + 0.5 )^2 - 12.25 = 0
Which is the same with (( x + 0.5 ) - 3.5 ) * (( x + 0.5 ) + 3.5 ) = 0
And it is the same with ( x + 0.5 - 3.5 ) * ( x + 0.5 + 3.5 ) = 0
Just add up the constants in each brackets, and we get ( x - 3 ) * ( x + 4 ) = 0
So we have the answers x1 = 3 and x2 = -4


Question Number 2 :
For this equation 6*x^2 - x - 3 = 0 , answer the following questions :
A. Find the roots using Quadratic Formula !
B. Use completing the square to find the root of the equation !

Answer Number 2 :
The equation 6*x^2 - x - 3 = 0 is already in a*x^2+b*x+c=0 form.
By matching the constant position, we can derive that the value of a = 6, b = -1, c = -3.

2A. Find the roots using Quadratic Formula !
By using abc formula the value of x is both
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
As we know that a = 6, b = -1 and c = -3,
we need to subtitute a,b,c in the abc formula, with thos values.
Which produce x1 = (-(-1) + sqrt( (-1)^2 - 4 * (6)*(-3)))/(2*6) and x2 = (-(-1) - sqrt( (-1)^2 - 4 * (6)*(-3)))/(2*6)
Which is the same as x1 = ( 1 + sqrt( 1+72))/(12) and x2 = ( 1 - sqrt( 1+72))/(12)
Which make x1 = ( 1 + sqrt( 73))/(12) and x2 = ( 1 - sqrt( 73))/(12)
We can get x1 = ( 1 + 8.54400374531753 )/(12) and x2 = ( 1 - 8.54400374531753 )/(12)
So we have the answers x1 = 0.795333645443128 and x2 = -0.628666978776461

2B. Use completing the square to find the root of the equation !
6*x^2 - x - 3 = 0 ,divide both side with 6
Which result in x^2 - 0.166666666666667*x - 0.5 = 0 ,
And the coefficient of x is -0.166666666666667
We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = -0.166666666666667/2 = -0.0833333333333333
Next, we have to separate the constant to form x^2 - 0.166666666666667*x + 0.00694444444444444 - 0.506944444444444 = 0
Which is the same with ( x - 0.0833333333333333 )^2 - 0.506944444444444 = 0
Which can be turned into (( x - 0.0833333333333333 ) - 0.712000312109794 ) * (( x - 0.0833333333333333 ) + 0.712000312109794 ) = 0
By using the associative law we get ( x - 0.0833333333333333 - 0.712000312109794 ) * ( x - 0.0833333333333333 + 0.712000312109794 ) = 0
Do the addition/subtraction, and we get ( x - 0.795333645443128 ) * ( x + 0.628666978776461 ) = 0
The answers are x1 = 0.795333645443128 and x2 = -0.628666978776461

For any 2nd degree or higher equation, set 1 side = 0. Factor, set factors = 0.

x^2 + x - 12 = 0
(x + 4)(x - 3) = 0
x + 4 = 0, x - 3 = 0
x = -4 or x = 3

x^2+x=12

subtract 12 from both sides

x^2+x-12 = 0

Factor

(x + 4)(x - 3) = 0

x = 3, -4
.

x^2+x=12
x^2+x-12=0

(x+4)(x-3)=0
x=-4 or 3

x^2 + x = 12
x^2 + x - 12 = 0
x^2 + 4x - 3x - 12 = 0
(x^2 + 4x) - (3x + 12) = 0
x(x + 4) - 3(x + 4) = 0
(x + 4)(x - 3) = 0

x + 4 = 0
x = -4

x - 3 = 0
x = 3

∴ x = -4, 3

x^2+x-12=0
x's=[-1+&-(1+48)^1/2]/2=3 & -4
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When,ax^2+bx+c=0
roots=x's=[-b+&-(b^2-4ac)^1/2] / 2a