2x^2+8x=20

Solution for 2x^2+8x=20 equation:

2x^2+8x=20

We move all terms to the left:

2x^2+8x-(20)=0

a = 2; b = 8; c = -20;
Δ = b2-4ac
Δ = 82-4·2·(-20)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{14}}{2*2}=\frac{-8-4\sqrt{14}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{14}}{2*2}=\frac{-8+4\sqrt{14}}{4}
$