m^2+8=80

Solution for m^2+8=80 equation:

m^2+8=80

We move all terms to the left:

m^2+8-(80)=0

We add all the numbers together, and all the variables

m^2-72=0

a = 1; b = 0; c = -72;
Δ = b2-4ac
Δ = 02-4·1·(-72)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*1}=\frac{0-12\sqrt{2}}{2}
=-\frac{12\sqrt{2}}{2}
=-6\sqrt{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*1}=\frac{0+12\sqrt{2}}{2}
=\frac{12\sqrt{2}}{2}
=6\sqrt{2}
$