2m^2-98=2

Solution for 2m^2-98=2 equation:

2m^2-98=2

We move all terms to the left:

2m^2-98-(2)=0

We add all the numbers together, and all the variables

2m^2-100=0

a = 2; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·2·(-100)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*2}=\frac{0-20\sqrt{2}}{4}
=-\frac{20\sqrt{2}}{4}
=-5\sqrt{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*2}=\frac{0+20\sqrt{2}}{4}
=\frac{20\sqrt{2}}{4}
=5\sqrt{2}
$