(x-9)(x+7)+(x+2)(x-3)=0

Solution for (x-9)(x+7)+(x+2)(x-3)=0 equation:

(x-9)(x+7)+(x+2)(x-3)=0

We multiply parentheses ..

(+x^2+7x-9x-63)+(x+2)(x-3)=0

We get rid of parentheses

x^2+7x-9x+(x+2)(x-3)-63=0

We multiply parentheses ..

x^2+(+x^2-3x+2x-6)+7x-9x-63=0

We add all the numbers together, and all the variables

x^2+(+x^2-3x+2x-6)-2x-63=0

We get rid of parentheses

x^2+x^2-3x+2x-2x-6-63=0

We add all the numbers together, and all the variables

2x^2-3x-69=0

a = 2; b = -3; c = -69;
Δ = b2-4ac
Δ = -32-4·2·(-69)
Δ = 561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{561}}{2*2}=\frac{3-\sqrt{561}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{561}}{2*2}=\frac{3+\sqrt{561}}{4}
$