(3x+4)(3x-4)=-(x-1)(x-10)

Solution for (3x+4)(3x-4)=-(x-1)(x-10) equation:

(3x+4)(3x-4)=-(x-1)(x-10)

We move all terms to the left:

(3x+4)(3x-4)-(-(x-1)(x-10))=0

We use the square of the difference formula

9x^2-(-(x-1)(x-10))-16=0

We multiply parentheses ..

9x^2-(-(+x^2-10x-1x+10))-16=0


We calculate terms in parentheses: -(-(+x^2-10x-1x+10)), so:

-(+x^2-10x-1x+10)

We get rid of parentheses

-x^2+10x+1x-10

We add all the numbers together, and all the variables

-1x^2+11x-10

Back to the equation:

-(-1x^2+11x-10)


We get rid of parentheses

9x^2+1x^2-11x+10-16=0

We add all the numbers together, and all the variables

10x^2-11x-6=0

a = 10; b = -11; c = -6;
Δ = b2-4ac
Δ = -112-4·10·(-6)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*10}=\frac{-8}{20}
=-2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*10}=\frac{30}{20}
=1+1/2
$