(9-3x)*(5+2x)+(x-3)*(5+x)=0

Solution for (9-3x)*(5+2x)+(x-3)*(5+x)=0 equation:

(9-3x)(5+2x)+(x-3)(5+x)=0

We add all the numbers together, and all the variables

(-3x+9)(2x+5)+(x-3)(x+5)=0

We multiply parentheses ..

(-6x^2-15x+18x+45)+(x-3)(x+5)=0

We get rid of parentheses

-6x^2-15x+18x+(x-3)(x+5)+45=0

We multiply parentheses ..

-6x^2+(+x^2+5x-3x-15)-15x+18x+45=0

We add all the numbers together, and all the variables

-6x^2+(+x^2+5x-3x-15)+3x+45=0

We get rid of parentheses

-6x^2+x^2+5x-3x+3x-15+45=0

We add all the numbers together, and all the variables

-5x^2+5x+30=0

a = -5; b = 5; c = +30;
Δ = b2-4ac
Δ = 52-4·(-5)·30
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-25}{2*-5}=\frac{-30}{-10}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+25}{2*-5}=\frac{20}{-10}
=-2
$