2x^2+20x=78

Solution for 2x^2+20x=78 equation:

2x^2+20x=78

We move all terms to the left:

2x^2+20x-(78)=0

a = 2; b = 20; c = -78;
Δ = b2-4ac
Δ = 202-4·2·(-78)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-32}{2*2}=\frac{-52}{4}
=-13
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+32}{2*2}=\frac{12}{4}
=3
$