4v^2+12=41

Solution for 4v^2+12=41 equation:

4v^2+12=41

We move all terms to the left:

4v^2+12-(41)=0

We add all the numbers together, and all the variables

4v^2-29=0

a = 4; b = 0; c = -29;
Δ = b2-4ac
Δ = 02-4·4·(-29)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{29}}{2*4}=\frac{0-4\sqrt{29}}{8}
=-\frac{4\sqrt{29}}{8}
=-\frac{\sqrt{29}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{29}}{2*4}=\frac{0+4\sqrt{29}}{8}
=\frac{4\sqrt{29}}{8}
=\frac{\sqrt{29}}{2}
$