-(3x-5)(x+4)=x(3x-7)+8

Solution for -(3x-5)(x+4)=x(3x-7)+8 equation:

-(3x-5)(x+4)=x(3x-7)+8

We move all terms to the left:

-(3x-5)(x+4)-(x(3x-7)+8)=0

We multiply parentheses ..

-(+3x^2+12x-5x-20)-(x(3x-7)+8)=0


We calculate terms in parentheses: -(x(3x-7)+8), so:

x(3x-7)+8

We multiply parentheses

3x^2-7x+8

Back to the equation:

-(3x^2-7x+8)


We get rid of parentheses

-3x^2-3x^2-12x+5x+7x+20-8=0

We add all the numbers together, and all the variables

-6x^2+12=0

a = -6; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-6)·12
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*-6}=\frac{0-12\sqrt{2}}{-12}
=-\frac{12\sqrt{2}}{-12}
=-\frac{\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*-6}=\frac{0+12\sqrt{2}}{-12}
=\frac{12\sqrt{2}}{-12}
=\frac{\sqrt{2}}{-1}
$