4n^2=29

Solution for 4n^2=29 equation:

4n^2=29

We move all terms to the left:

4n^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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$n_{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}
$
$n_{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}
$