(2x+10)(2x+20)=400

Solution for (2x+10)(2x+20)=400 equation:

(2x+10)(2x+20)=400

We move all terms to the left:

(2x+10)(2x+20)-(400)=0

We multiply parentheses ..

(+4x^2+40x+20x+200)-400=0

We get rid of parentheses

4x^2+40x+20x+200-400=0

We add all the numbers together, and all the variables

4x^2+60x-200=0

a = 4; b = 60; c = -200;
Δ = b2-4ac
Δ = 602-4·4·(-200)
Δ = 6800
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:

$\sqrt{\Delta}=\sqrt{6800}=\sqrt{400*17}=\sqrt{400}*\sqrt{17}=20\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-20\sqrt{17}}{2*4}=\frac{-60-20\sqrt{17}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+20\sqrt{17}}{2*4}=\frac{-60+20\sqrt{17}}{8}
$