(x-3^2)+(x+6)(x-3)=5

Solution for (x-3^2)+(x+6)(x-3)=5 equation:

(x-3^2)+(x+6)(x-3)=5

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

(+x^2-3x+6x-18)+x-14=0

We get rid of parentheses

x^2-3x+6x+x-18-14=0

We add all the numbers together, and all the variables

x^2+4x-32=0

a = 1; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·1·(-32)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*1}=\frac{8}{2}
=4
$