3n2-25=28

Solution for 3n2-25=28 equation:

3n^2-25=28

We move all terms to the left:

3n^2-25-(28)=0

We add all the numbers together, and all the variables

3n^2-53=0

a = 3; b = 0; c = -53;
Δ = b2-4ac
Δ = 02-4·3·(-53)
Δ = 636
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:

$\sqrt{\Delta}=\sqrt{636}=\sqrt{4*159}=\sqrt{4}*\sqrt{159}=2\sqrt{159}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{159}}{2*3}=\frac{0-2\sqrt{159}}{6}
=-\frac{2\sqrt{159}}{6}
=-\frac{\sqrt{159}}{3}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{159}}{2*3}=\frac{0+2\sqrt{159}}{6}
=\frac{2\sqrt{159}}{6}
=\frac{\sqrt{159}}{3}
$