(x^2+3x)+(2-15x)=90

Solution for (x^2+3x)+(2-15x)=90 equation:

(x^2+3x)+(2-15x)=90

We move all terms to the left:

(x^2+3x)+(2-15x)-(90)=0

We add all the numbers together, and all the variables

(x^2+3x)+(-15x+2)-90=0

We get rid of parentheses

x^2+3x-15x+2-90=0

We add all the numbers together, and all the variables

x^2-12x-88=0

a = 1; b = -12; c = -88;
Δ = b2-4ac
Δ = -122-4·1·(-88)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{31}}{2*1}=\frac{12-4\sqrt{31}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{31}}{2*1}=\frac{12+4\sqrt{31}}{2}
$