v^2+4=67

Solution for v^2+4=67 equation:

v^2+4=67

We move all terms to the left:

v^2+4-(67)=0

We add all the numbers together, and all the variables

v^2-63=0

a = 1; b = 0; c = -63;
Δ = b2-4ac
Δ = 02-4·1·(-63)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{7}}{2*1}=\frac{0-6\sqrt{7}}{2}
=-\frac{6\sqrt{7}}{2}
=-3\sqrt{7}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{7}}{2*1}=\frac{0+6\sqrt{7}}{2}
=\frac{6\sqrt{7}}{2}
=3\sqrt{7}
$