w+w+w^2-4+w^2-3=60

Solution for w+w+w^2-4+w^2-3=60 equation:

w+w+w^2-4+w^2-3=60

We move all terms to the left:

w+w+w^2-4+w^2-3-(60)=0

We add all the numbers together, and all the variables

2w^2+2w-67=0

a = 2; b = 2; c = -67;
Δ = b2-4ac
Δ = 22-4·2·(-67)
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{15}}{2*2}=\frac{-2-6\sqrt{15}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{15}}{2*2}=\frac{-2+6\sqrt{15}}{4}
$