2x^2+40x=138

Solution for 2x^2+40x=138 equation:

2x^2+40x=138

We move all terms to the left:

2x^2+40x-(138)=0

a = 2; b = 40; c = -138;
Δ = b2-4ac
Δ = 402-4·2·(-138)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-52}{2*2}=\frac{-92}{4}
=-23
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+52}{2*2}=\frac{12}{4}
=3
$