3w+w^2-90=0

Solution for 3w+w^2-90=0 equation:

3w+w^2-90=0

a = 1; b = 3; c = -90;
Δ = b2-4ac
Δ = 32-4·1·(-90)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{41}}{2*1}=\frac{-3-3\sqrt{41}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{41}}{2*1}=\frac{-3+3\sqrt{41}}{2}
$