(2x+5)(4x-10)+65=180

Solution for (2x+5)(4x-10)+65=180 equation:

(2x+5)(4x-10)+65=180

We move all terms to the left:

(2x+5)(4x-10)+65-(180)=0

We add all the numbers together, and all the variables

(2x+5)(4x-10)-115=0

We multiply parentheses ..

(+8x^2-20x+20x-50)-115=0

We get rid of parentheses

8x^2-20x+20x-50-115=0

We add all the numbers together, and all the variables

8x^2-165=0

a = 8; b = 0; c = -165;
Δ = b2-4ac
Δ = 02-4·8·(-165)
Δ = 5280
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:

$\sqrt{\Delta}=\sqrt{5280}=\sqrt{16*330}=\sqrt{16}*\sqrt{330}=4\sqrt{330}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{330}}{2*8}=\frac{0-4\sqrt{330}}{16}
=-\frac{4\sqrt{330}}{16}
=-\frac{\sqrt{330}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{330}}{2*8}=\frac{0+4\sqrt{330}}{16}
=\frac{4\sqrt{330}}{16}
=\frac{\sqrt{330}}{4}
$