20x^2+80x+50=2000

Solution for 20x^2+80x+50=2000 equation:

20x^2+80x+50=2000

We move all terms to the left:

20x^2+80x+50-(2000)=0

We add all the numbers together, and all the variables

20x^2+80x-1950=0

a = 20; b = 80; c = -1950;
Δ = b2-4ac
Δ = 802-4·20·(-1950)
Δ = 162400
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:

$\sqrt{\Delta}=\sqrt{162400}=\sqrt{400*406}=\sqrt{400}*\sqrt{406}=20\sqrt{406}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-20\sqrt{406}}{2*20}=\frac{-80-20\sqrt{406}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+20\sqrt{406}}{2*20}=\frac{-80+20\sqrt{406}}{40}
$