solve quadratic formula x^2-10x-1=-10?

1) factorise the quadratic equation

x^2 - 10x - 1 = -10
x^2 - 10x +9 = 0
(x-9)(x-1)= 0

2) Apply the Null Factor Law

Either x-9 = 0 or x - 1 = 0
x = 9 or x = 1
Solution: (x : 1, 9)

x² - 10x + 9 = 0
(x - 9)(x - 1) = 0
x = 9 , x = 1
OR
x = [ 10 ± √ (100 - 36) ] / 2
x = [ 10 ± √ (64) ] / 2
x = [ 10 ± 8 ] / 2
x = 9 , x = 1

Question Number 1 :
For this equation x^2 - 10*x - 1 = - 10 , answer the following questions :
A. Calculate the Roots ( x1 and x2 ) !
B. Use factorization to find the root of the equation !

Answer Number 1 :
First, we have to turn equation : x^2 - 10*x - 1 = - 10 , into a*x^2+b*x+c=0 form.
x^2 - 10*x - 1 = - 10 , move everything in the right hand side, to the left hand side of the equation
<=> x^2 - 10*x - 1 - ( - 10 ) = 0 , which is the same with
<=> x^2 - 10*x - 1 + ( 10 ) =0 , now open the bracket and we get
<=> x^2 - 10*x + 9 = 0

The equation x^2 - 10*x + 9 = 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 = -10, c = 9.

1A. Calculate the Roots ( x1 and x2 ) !
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 a = 1, b = -10 and c = 9,
we just need to subtitute the value of a,b and c in the abc formula.
Which produce x1 = (-(-10) + sqrt( (-10)^2 - 4 * (1)*(9)))/(2*1) and x2 = (-(-10) - sqrt( (-10)^2 - 4 * (1)*(9)))/(2*1)
Which is can be turned into x1 = ( 10 + sqrt( 100-36))/(2) and x2 = ( 10 - sqrt( 100-36))/(2)
Which is the same as x1 = ( 10 + sqrt( 64))/(2) and x2 = ( 10 - sqrt( 64))/(2)
We can get x1 = ( 10 + 8 )/(2) and x2 = ( 10 - 8 )/(2)
We get following answers x1 = 9 and x2 = 1

1B. Use factorization to find the root of the equation !
x^2 - 10*x + 9 = 0
<=> ( x - 9 ) * ( x - 1 ) = 0
The answers are x1 = 9 and x2 = 1

Your answer doesn't seems right.

x^2 - 10x - 1 = -10
x^2 - 10x - 1 + 10 = 0
x^2 - 10x + 9 = 0
(x - 9)(x - 1) = 0

x - 9 = 0
x = 9

x - 1 = 0
x = 1

∴ x = 9 , 1

x^2-10x+9=0
a=1
b=-10
c=9
the formula is under= [-b(+-) root(b^2-4ac)]/2a
so root(b^2-4ac) this comes out to under root of(100-(4*1*9)) which is 8

(-b+8)/2*1 ---------put b=-10
(10+8)/2=9
similarly
(-b-8)/2*1-------------put b=-10
(10-8)/2=1
so
the ans. are 1and 9

you made a mistake. the solution goes this way:
Formula for quadratic equation is

[-b+-square root{b^2-4.a.c}]/2.a
here, a= 1, b= -10, c= 9
if you solve it, you get [10+-8]/2
= 9 or 1

next way to solve this is
x^2-10x+9=0
or, x^2-x-9x+9=0
or, x(x-1)-9(x-1)=0
or, (x-1) (x-9)=o
so, x= 1 or 9

Please, be more specific.

I will find x for you lol.

x^2-10x-1=-10
x^2-10x+9=0

Where a=1 b=-10 c=9

b^2-4ac= (-10)^2-4(1)(9) = 100 - 36 = 64

Square root of 64 is 8

The formula to find x is:
x= [-b + or - squarerootb^2-4ac] / 2a

x'= [-(-10) + 8] / 2 = ( 10 + 8 )/2 = 9
x'= [-(-10) - 8] / 2 = ( 10 - 8 )/2 = 1

So x = 9 or 1

Then, you just need to replace

(9)^2-10(9)-1=-10
81-90-1 = -10
-9 -1 = -10
-10= -10

(1)^2-10(1)-1=-10
1-10-1=-10
-10=-10

The 2 points: (9,-10) and (1,-10)

You rearranged the equation ok, but the first term in the formula is -b, and the b under the square root is squared. Note that the coefficients (a, b, c) have to include the sign, so b is actually -10. So you should have:
x=(10+-sqrt(100-36))/2
=(10+-sqrt(64))/2
=(10+-8)/2
=9 and 1

Substituting in the original equation:
(9)^2-10(9)-1=-10
81-90-1=-10
-10=-10

(1)^2-10(1)-1=-10
1-10-1=-10
-10=-10

don't ever leave a square root on the bottom of a fraction. if you're dealing with a quadratic equation first you want everything on the left and zero on the right.

X^2-10x-1=-10
+10 +10
X^2-10x+9=0

then you can factor into parentheses. you need two numbers that multiply to +9 and add to -10: -1 and -9

(x-9)(x-1)=0

set each parentheses = to 0
x-9=0 and x-1=0
x=9 x=1