3w(w-5)+16=0

Solution for 3w(w-5)+16=0 equation:

3w(w-5)+16=0

We multiply parentheses

3w^2-15w+16=0

a = 3; b = -15; c = +16;
Δ = b2-4ac
Δ = -152-4·3·16
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{33}}{2*3}=\frac{15-\sqrt{33}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{33}}{2*3}=\frac{15+\sqrt{33}}{6}
$