2x^2+8x=1020

Solution for 2x^2+8x=1020 equation:

2x^2+8x=1020

We move all terms to the left:

2x^2+8x-(1020)=0

a = 2; b = 8; c = -1020;
Δ = b2-4ac
Δ = 82-4·2·(-1020)
Δ = 8224
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:

$\sqrt{\Delta}=\sqrt{8224}=\sqrt{16*514}=\sqrt{16}*\sqrt{514}=4\sqrt{514}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{514}}{2*2}=\frac{-8-4\sqrt{514}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{514}}{2*2}=\frac{-8+4\sqrt{514}}{4}
$