3w^2+6w-15=0

Solution for 3w^2+6w-15=0 equation:

3w^2+6w-15=0

a = 3; b = 6; c = -15;
Δ = b2-4ac
Δ = 62-4·3·(-15)
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{6}}{2*3}=\frac{-6-6\sqrt{6}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{6}}{2*3}=\frac{-6+6\sqrt{6}}{6}
$