.3u^2+15u-3=0

Solution for .3u^2+15u-3=0 equation:

.3u^2+15u-3=0

a = .3; b = 15; c = -3;
Δ = b2-4ac
Δ = 152-4·.3·(-3)
Δ = 228.6
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{228.6}}{2*.3}=\frac{-15-\sqrt{228.6}}{0.6}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{228.6}}{2*.3}=\frac{-15+\sqrt{228.6}}{0.6}
$