3x^2+4x=118

Solution for 3x^2+4x=118 equation:

3x^2+4x=118

We move all terms to the left:

3x^2+4x-(118)=0

a = 3; b = 4; c = -118;
Δ = b2-4ac
Δ = 42-4·3·(-118)
Δ = 1432
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:

$\sqrt{\Delta}=\sqrt{1432}=\sqrt{4*358}=\sqrt{4}*\sqrt{358}=2\sqrt{358}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{358}}{2*3}=\frac{-4-2\sqrt{358}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{358}}{2*3}=\frac{-4+2\sqrt{358}}{6}
$