(x-4)(x+3)=6x+2x2

Solution for (x-4)(x+3)=6x+2x2 equation:

(x-4)(x+3)=6x+2x^2

We move all terms to the left:

(x-4)(x+3)-(6x+2x^2)=0

We get rid of parentheses

-2x^2-6x+(x-4)(x+3)=0

We multiply parentheses ..

-2x^2+(+x^2+3x-4x-12)-6x=0

We get rid of parentheses

-2x^2+x^2+3x-4x-6x-12=0

We add all the numbers together, and all the variables

-1x^2-7x-12=0

a = -1; b = -7; c = -12;
Δ = b2-4ac
Δ = -72-4·(-1)·(-12)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*-1}=\frac{6}{-2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*-1}=\frac{8}{-2}
=-4
$