m^2-32=18

Solution for m^2-32=18 equation:

m^2-32=18

We move all terms to the left:

m^2-32-(18)=0

We add all the numbers together, and all the variables

m^2-50=0

a = 1; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·1·(-50)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*1}=\frac{0-10\sqrt{2}}{2}
=-\frac{10\sqrt{2}}{2}
=-5\sqrt{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*1}=\frac{0+10\sqrt{2}}{2}
=\frac{10\sqrt{2}}{2}
=5\sqrt{2}
$