(x-2)(x+5)=(x-3)(x+4)+x2

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Solution for (x-2)(x+5)=(x-3)(x+4)+x2 equation:



(x-2)(x+5)=(x-3)(x+4)+x2
We move all terms to the left:
(x-2)(x+5)-((x-3)(x+4)+x2)=0
We multiply parentheses ..
(+x^2+5x-2x-10)-((x-3)(x+4)+x2)=0
We calculate terms in parentheses: -((x-3)(x+4)+x2), so:
(x-3)(x+4)+x2
We add all the numbers together, and all the variables
x^2+(x-3)(x+4)
We multiply parentheses ..
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
x^2-2x^2+5x-2x-x-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} $

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