(2x+10)(2x+20)=216

Solution for (2x+10)(2x+20)=216 equation:

(2x+10)(2x+20)=216

We move all terms to the left:

(2x+10)(2x+20)-(216)=0

We multiply parentheses ..

(+4x^2+40x+20x+200)-216=0

We get rid of parentheses

4x^2+40x+20x+200-216=0

We add all the numbers together, and all the variables

4x^2+60x-16=0

a = 4; b = 60; c = -16;
Δ = b2-4ac
Δ = 602-4·4·(-16)
Δ = 3856
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:

$\sqrt{\Delta}=\sqrt{3856}=\sqrt{16*241}=\sqrt{16}*\sqrt{241}=4\sqrt{241}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-4\sqrt{241}}{2*4}=\frac{-60-4\sqrt{241}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+4\sqrt{241}}{2*4}=\frac{-60+4\sqrt{241}}{8}
$