2x(x-5)=10x+200

Solution for 2x(x-5)=10x+200 equation:

2x(x-5)=10x+200

We move all terms to the left:

2x(x-5)-(10x+200)=0

We multiply parentheses

2x^2-10x-(10x+200)=0

We get rid of parentheses

2x^2-10x-10x-200=0

We add all the numbers together, and all the variables

2x^2-20x-200=0

a = 2; b = -20; c = -200;
Δ = b2-4ac
Δ = -202-4·2·(-200)
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{5}}{2*2}=\frac{20-20\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{5}}{2*2}=\frac{20+20\sqrt{5}}{4}
$