(x-2)(x+5)=(x-3)(x+4)+x^2

Solution for (x-2)(x+5)=(x-3)(x+4)+x^2 equation:

(x-2)(x+5)=(x-3)(x+4)+x^2

We move all terms to the left:

(x-2)(x+5)-((x-3)(x+4)+x^2)=0

We multiply parentheses ..

-((+x^2+4x-3x-12)+x^2)+(x-2)(x+5)=0


We calculate terms in parentheses: -((+x^2+4x-3x-12)+x^2), so:

(+x^2+4x-3x-12)+x^2

We add all the numbers together, and all the variables

x^2+(+x^2+4x-3x-12)

We get rid of parentheses

x^2+x^2+4x-3x-12

We add all the numbers together, and all the variables

2x^2+x-12

Back to the equation:

-(2x^2+x-12)


We get rid of parentheses

-2x^2-x+(x-2)(x+5)+12=0

We multiply parentheses ..

-2x^2+(+x^2+5x-2x-10)-x+12=0

We add all the numbers together, and all the variables

-2x^2+(+x^2+5x-2x-10)-1x+12=0

We get rid of parentheses

-2x^2+x^2+5x-2x-1x-10+12=0

We add all the numbers together, and all the variables

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

a = -1; b = 2; c = +2;
Δ = b2-4ac
Δ = 22-4·(-1)·2
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3}}{2*-1}=\frac{-2-2\sqrt{3}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3}}{2*-1}=\frac{-2+2\sqrt{3}}{-2}
$