3x^2+4x=320

Solution for 3x^2+4x=320 equation:

3x^2+4x=320

We move all terms to the left:

3x^2+4x-(320)=0

a = 3; b = 4; c = -320;
Δ = b2-4ac
Δ = 42-4·3·(-320)
Δ = 3856
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:

$\sqrt{\Delta}=\sqrt{3856}=\sqrt{16*241}=\sqrt{16}*\sqrt{241}=4\sqrt{241}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{241}}{2*3}=\frac{-4-4\sqrt{241}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{241}}{2*3}=\frac{-4+4\sqrt{241}}{6}
$