180=(2x+4)(2x-9)

Solution for 180=(2x+4)(2x-9) equation:

180=(2x+4)(2x-9)

We move all terms to the left:

180-((2x+4)(2x-9))=0

We multiply parentheses ..

-((+4x^2-18x+8x-36))+180=0


We calculate terms in parentheses: -((+4x^2-18x+8x-36)), so:

(+4x^2-18x+8x-36)

We get rid of parentheses

4x^2-18x+8x-36

We add all the numbers together, and all the variables

4x^2-10x-36

Back to the equation:

-(4x^2-10x-36)


We get rid of parentheses

-4x^2+10x+36+180=0

We add all the numbers together, and all the variables

-4x^2+10x+216=0

a = -4; b = 10; c = +216;
Δ = b2-4ac
Δ = 102-4·(-4)·216
Δ = 3556
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:

$\sqrt{\Delta}=\sqrt{3556}=\sqrt{4*889}=\sqrt{4}*\sqrt{889}=2\sqrt{889}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{889}}{2*-4}=\frac{-10-2\sqrt{889}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{889}}{2*-4}=\frac{-10+2\sqrt{889}}{-8}
$