v^2+4=63

Solution for v^2+4=63 equation:

v^2+4=63

We move all terms to the left:

v^2+4-(63)=0

We add all the numbers together, and all the variables

v^2-59=0

a = 1; b = 0; c = -59;
Δ = b2-4ac
Δ = 02-4·1·(-59)
Δ = 236
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{236}=\sqrt{4*59}=\sqrt{4}*\sqrt{59}=2\sqrt{59}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{59}}{2*1}=\frac{0-2\sqrt{59}}{2}
=-\frac{2\sqrt{59}}{2}
=-\sqrt{59}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{59}}{2*1}=\frac{0+2\sqrt{59}}{2}
=\frac{2\sqrt{59}}{2}
=\sqrt{59}
$