2x2+40x=80

Solution for 2x2+40x=80 equation:

2x^2+40x=80

We move all terms to the left:

2x^2+40x-(80)=0

a = 2; b = 40; c = -80;
Δ = b2-4ac
Δ = 402-4·2·(-80)
Δ = 2240
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:

$\sqrt{\Delta}=\sqrt{2240}=\sqrt{64*35}=\sqrt{64}*\sqrt{35}=8\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{35}}{2*2}=\frac{-40-8\sqrt{35}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{35}}{2*2}=\frac{-40+8\sqrt{35}}{4}
$