. Solve using the quadratic formula:?

The quadradic formula is:

x = [ -b ± √(b² - 4ac)] / (2a)

You have to get the formula into the following format for it to work:

ax² + bx + c = 0

So first, get it in the format:

x² - 7x - 1 = -7
x² - 7x + 6 = 0

Now we have:

a = 1
b = -7
c = 6

Plug in what we know into the formula:

x = [ -b ± √(b² - 4ac)] / (2a)
x = [ -(-7) ± √((-7)² - 4(1)(6))] / (2(1))
x = [ 7 ± √(49 - 24)] / (2)
x = [ 7 ± √(25)] / (2)
x = [ 7 ± 5 ] / (2)

x = 12 / 2 and x = 2/2
x = 6 and 1

question huge style a million : For this equation x^2 - 7*x - a million = - 7 , answer here questions : A. discover the roots using Quadratic formula ! B. Use factorization to discover the basis of the equation ! C. Use winding up the sq. to discover the basis of the equation ! answer huge style a million : First, we could desire to tutor equation : x^2 - 7*x - a million = - 7 , right into a*x^2+b*x+c=0 sort. x^2 - 7*x - a million = - 7 , pass each and every thing in the terrific hand edge, to the left hand edge of the equation <=> x^2 - 7*x - a million - ( - 7 ) = 0 , it fairly is the comparable with <=> x^2 - 7*x - a million + ( 7 ) =0 , now open the bracket and we get <=> x^2 - 7*x + 6 = 0 The equation x^2 - 7*x + 6 = 0 is already in a*x^2+b*x+c=0 sort. In that kind, we can definitely derive that the linked fee of a = a million, b = -7, c = 6. 1A. discover the roots using Quadratic formula ! Use the formula, x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a) We had understand that a = a million, b = -7 and c = 6, we could desire to subtitute a,b,c in the abc formula, with thos values. Which produce x1 = (-(-7) + sqrt( (-7)^2 - 4 * (a million)*(6)))/(2*a million) and x2 = (-(-7) - sqrt( (-7)^2 - 4 * (a million)*(6)))/(2*a million) it fairly is the comparable with x1 = ( 7 + sqrt( 40 9-24))/(2) and x2 = ( 7 - sqrt( 40 9-24))/(2) Which make x1 = ( 7 + sqrt( 25))/(2) and x2 = ( 7 - sqrt( 25))/(2) So we get x1 = ( 7 + 5 )/(2) and x2 = ( 7 - 5 )/(2) So we've the solutions x1 = 6 and x2 = a million 1B. Use factorization to discover the basis of the equation ! x^2 - 7*x + 6 = 0 ( x - 6 ) * ( x - a million ) = 0 The solutions are x1 = 6 and x2 = a million 1C. Use winding up the sq. to discover the basis of the equation ! x^2 - 7*x + 6 = 0 ,divide the two edge with a million Then we get x^2 - 7*x + 6 = 0 , all of us understand that the coefficient of x is -7 we could desire to apply the undeniable fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = -7/2 = -3.5 So we've make the equation into x^2 - 7*x + 12.25 - 6.25 = 0 which would be became into ( x - 3.5 )^2 - 6.25 = 0 So we can get (( x - 3.5 ) - 2.5 ) * (( x - 3.5 ) + 2.5 ) = 0 via using the associative regulation we get ( x - 3.5 - 2.5 ) * ( x - 3.5 + 2.5 ) = 0 And it fairly is the comparable with ( x - 6 ) * ( x - a million ) = 0 So we've been given the solutions as x1 = 6 and x2 = a million

x² - 7x + 6 = 0
(x - 6)(x - 1) = 0
x = 6 , x = 1

OR

x = [ - b ± √ (b ² - 4 a c ) ] / 2 a
x = [ 7 ± √ (49 - 24 ) ] / 2
x = [ 7 ± √ (25) ] / 2
x = [ 7 ± 5 ] / 2
x = 6 , x = 1

The Quatratic equation is solved from the following format:
Ax^2 +Bx+C=0

so, first, lets get the equation written in that form:
x^2-7X-1 = -7
x^2-7X+6 = 0
A=1, B=-7, C=6

Quadratic formula states:
x= (-b +/- sqrt(b^2-4AC))/2A
= (-b + sqrt(b^2-4AC))/2A
= (-b - sqrt(b^2-4AC))/2A
= (7 + sqrt(7^2-4*1*6))/2*1
= (7 - sqrt(7^2-4*1*6))/2*1
= (7 + sqrt(49-24))/2
= (7 - sqrt(49-24))/2
= (7 + sqrt(25))/2
= (7 - sqrt(25))/2
= (7 + 5)/2
= (7 - 5)/2
= (2)/2
= (12)/2
=6 or 1

x^2 - 7x + 6 = 0
x = (-b +/- sqrt(b^2 - 4ac))/2a
x = (7 +/- sqrt(49 - 24))/2
x = (7 +/- sqrt(25))/2
x = (7 +/- 5)/2
x = 12/2, 2/2
x = 6, 1

x^2 - 7x - 1 = -7
x^2 - 7x - 1 + 7 = 0
x^2 - 7x + 6 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = -7
c = 6

x = [7 ±√(49 - 24)]/2
x = [7 ±√25]/2
x = [7 ±5]/2

x = [7 + 5]/2
x = 12/2
x = 6

x = [7 - 5]/2
x = 2/2
x = 1

∴ x = 1 , 6