4(a^2-62)=0

Solution for 4(a^2-62)=0 equation:

4(a^2-62)=0

We multiply parentheses

4a^2-248=0

a = 4; b = 0; c = -248;
Δ = b2-4ac
Δ = 02-4·4·(-248)
Δ = 3968
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3968}=\sqrt{64*62}=\sqrt{64}*\sqrt{62}=8\sqrt{62}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{62}}{2*4}=\frac{0-8\sqrt{62}}{8}
=-\frac{8\sqrt{62}}{8}
=-\sqrt{62}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{62}}{2*4}=\frac{0+8\sqrt{62}}{8}
=\frac{8\sqrt{62}}{8}
=\sqrt{62}
$