3u^2+8u-4=0

Solution for 3u^2+8u-4=0 equation:

3u^2+8u-4=0

a = 3; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·3·(-4)
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{7}}{2*3}=\frac{-8-4\sqrt{7}}{6}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{7}}{2*3}=\frac{-8+4\sqrt{7}}{6}
$