4x^2+40x=800

Solution for 4x^2+40x=800 equation:

4x^2+40x=800

We move all terms to the left:

4x^2+40x-(800)=0

a = 4; b = 40; c = -800;
Δ = b2-4ac
Δ = 402-4·4·(-800)
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-120}{2*4}=\frac{-160}{8}
=-20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+120}{2*4}=\frac{80}{8}
=10
$