90=(2x+2)(3x-5)

Solution for 90=(2x+2)(3x-5) equation:

90=(2x+2)(3x-5)

We move all terms to the left:

90-((2x+2)(3x-5))=0

We multiply parentheses ..

-((+6x^2-10x+6x-10))+90=0


We calculate terms in parentheses: -((+6x^2-10x+6x-10)), so:

(+6x^2-10x+6x-10)

We get rid of parentheses

6x^2-10x+6x-10

We add all the numbers together, and all the variables

6x^2-4x-10

Back to the equation:

-(6x^2-4x-10)


We get rid of parentheses

-6x^2+4x+10+90=0

We add all the numbers together, and all the variables

-6x^2+4x+100=0

a = -6; b = 4; c = +100;
Δ = b2-4ac
Δ = 42-4·(-6)·100
Δ = 2416
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:

$\sqrt{\Delta}=\sqrt{2416}=\sqrt{16*151}=\sqrt{16}*\sqrt{151}=4\sqrt{151}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{151}}{2*-6}=\frac{-4-4\sqrt{151}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{151}}{2*-6}=\frac{-4+4\sqrt{151}}{-12}
$