(X-4)(x+5)-(3x+2x^2-30)=0

Solution for (X-4)(x+5)-(3x+2x^2-30)=0 equation:

(X-4)(X+5)-(3X+2X^2-30)=0

We get rid of parentheses

-2X^2-3X+(X-4)(X+5)+30=0

We multiply parentheses ..

-2X^2+(+X^2+5X-4X-20)-3X+30=0

We get rid of parentheses

-2X^2+X^2+5X-4X-3X-20+30=0

We add all the numbers together, and all the variables

-1X^2-2X+10=0

a = -1; b = -2; c = +10;
Δ = b2-4ac
Δ = -22-4·(-1)·10
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{11}}{2*-1}=\frac{2-2\sqrt{11}}{-2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{11}}{2*-1}=\frac{2+2\sqrt{11}}{-2}
$