w^2+12=93

Solution for w^2+12=93 equation:

w^2+12=93

We move all terms to the left:

w^2+12-(93)=0

We add all the numbers together, and all the variables

w^2-81=0

a = 1; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·1·(-81)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*1}=\frac{-18}{2}
=-9
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*1}=\frac{18}{2}
=9
$