60w+4w^2=396

Solution for 60w+4w^2=396 equation:

60w+4w^2=396

We move all terms to the left:

60w+4w^2-(396)=0

a = 4; b = 60; c = -396;
Δ = b2-4ac
Δ = 602-4·4·(-396)
Δ = 9936
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:

$\sqrt{\Delta}=\sqrt{9936}=\sqrt{144*69}=\sqrt{144}*\sqrt{69}=12\sqrt{69}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-12\sqrt{69}}{2*4}=\frac{-60-12\sqrt{69}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+12\sqrt{69}}{2*4}=\frac{-60+12\sqrt{69}}{8}
$