v^2+1=96

Solution for v^2+1=96 equation:

v^2+1=96

We move all terms to the left:

v^2+1-(96)=0

We add all the numbers together, and all the variables

v^2-95=0

a = 1; b = 0; c = -95;
Δ = b2-4ac
Δ = 02-4·1·(-95)
Δ = 380
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:

$\sqrt{\Delta}=\sqrt{380}=\sqrt{4*95}=\sqrt{4}*\sqrt{95}=2\sqrt{95}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{95}}{2*1}=\frac{0-2\sqrt{95}}{2}
=-\frac{2\sqrt{95}}{2}
=-\sqrt{95}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{95}}{2*1}=\frac{0+2\sqrt{95}}{2}
=\frac{2\sqrt{95}}{2}
=\sqrt{95}
$