400=4x^2+80x-150

Solution for 400=4x^2+80x-150 equation:

400=4x^2+80x-150

We move all terms to the left:

400-(4x^2+80x-150)=0

We get rid of parentheses

-4x^2-80x+150+400=0

We add all the numbers together, and all the variables

-4x^2-80x+550=0

a = -4; b = -80; c = +550;
Δ = b2-4ac
Δ = -802-4·(-4)·550
Δ = 15200
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:

$\sqrt{\Delta}=\sqrt{15200}=\sqrt{400*38}=\sqrt{400}*\sqrt{38}=20\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-20\sqrt{38}}{2*-4}=\frac{80-20\sqrt{38}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+20\sqrt{38}}{2*-4}=\frac{80+20\sqrt{38}}{-8}
$