4n^2=280

Solution for 4n^2=280 equation:

4n^2=280

We move all terms to the left:

4n^2-(280)=0

a = 4; b = 0; c = -280;
Δ = b2-4ac
Δ = 02-4·4·(-280)
Δ = 4480
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:

$\sqrt{\Delta}=\sqrt{4480}=\sqrt{64*70}=\sqrt{64}*\sqrt{70}=8\sqrt{70}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{70}}{2*4}=\frac{0-8\sqrt{70}}{8}
=-\frac{8\sqrt{70}}{8}
=-\sqrt{70}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{70}}{2*4}=\frac{0+8\sqrt{70}}{8}
=\frac{8\sqrt{70}}{8}
=\sqrt{70}
$