m2+15=128

Solution for m2+15=128 equation:

m2+15=128

We move all terms to the left:

m2+15-(128)=0

We add all the numbers together, and all the variables

m^2-113=0

a = 1; b = 0; c = -113;
Δ = b2-4ac
Δ = 02-4·1·(-113)
Δ = 452
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{452}=\sqrt{4*113}=\sqrt{4}*\sqrt{113}=2\sqrt{113}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{113}}{2*1}=\frac{0-2\sqrt{113}}{2}
=-\frac{2\sqrt{113}}{2}
=-\sqrt{113}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{113}}{2*1}=\frac{0+2\sqrt{113}}{2}
=\frac{2\sqrt{113}}{2}
=\sqrt{113}
$