33x^2+40x=78

Solution for 33x^2+40x=78 equation:

33x^2+40x=78

We move all terms to the left:

33x^2+40x-(78)=0

a = 33; b = 40; c = -78;
Δ = b2-4ac
Δ = 402-4·33·(-78)
Δ = 11896
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:

$\sqrt{\Delta}=\sqrt{11896}=\sqrt{4*2974}=\sqrt{4}*\sqrt{2974}=2\sqrt{2974}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{2974}}{2*33}=\frac{-40-2\sqrt{2974}}{66}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{2974}}{2*33}=\frac{-40+2\sqrt{2974}}{66}
$