2x^2+192=40x

Solution for 2x^2+192=40x equation:

2x^2+192=40x

We move all terms to the left:

2x^2+192-(40x)=0

a = 2; b = -40; c = +192;
Δ = b2-4ac
Δ = -402-4·2·192
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8}{2*2}=\frac{32}{4}
=8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8}{2*2}=\frac{48}{4}
=12
$