3x^2+4x=112

Solution for 3x^2+4x=112 equation:

3x^2+4x=112

We move all terms to the left:

3x^2+4x-(112)=0

a = 3; b = 4; c = -112;
Δ = b2-4ac
Δ = 42-4·3·(-112)
Δ = 1360
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:

$\sqrt{\Delta}=\sqrt{1360}=\sqrt{16*85}=\sqrt{16}*\sqrt{85}=4\sqrt{85}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{85}}{2*3}=\frac{-4-4\sqrt{85}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{85}}{2*3}=\frac{-4+4\sqrt{85}}{6}
$