a^2-4=96

Solution for a^2-4=96 equation:

a^2-4=96

We move all terms to the left:

a^2-4-(96)=0

We add all the numbers together, and all the variables

a^2-100=0

a = 1; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·1·(-100)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*1}=\frac{-20}{2}
=-10
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*1}=\frac{20}{2}
=10
$