33w+90=3w^2

Solution for 33w+90=3w^2 equation:

33w+90=3w^2

We move all terms to the left:

33w+90-(3w^2)=0

determiningTheFunctionDomain
-3w^2+33w+90=0

a = -3; b = 33; c = +90;
Δ = b2-4ac
Δ = 332-4·(-3)·90
Δ = 2169
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:

$\sqrt{\Delta}=\sqrt{2169}=\sqrt{9*241}=\sqrt{9}*\sqrt{241}=3\sqrt{241}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{241}}{2*-3}=\frac{-33-3\sqrt{241}}{-6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{241}}{2*-3}=\frac{-33+3\sqrt{241}}{-6}
$