2m^2+10=200

Solution for 2m^2+10=200 equation:

2m^2+10=200

We move all terms to the left:

2m^2+10-(200)=0

We add all the numbers together, and all the variables

2m^2-190=0

a = 2; b = 0; c = -190;
Δ = b2-4ac
Δ = 02-4·2·(-190)
Δ = 1520
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:

$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{95}}{2*2}=\frac{0-4\sqrt{95}}{4}
=-\frac{4\sqrt{95}}{4}
=-\sqrt{95}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{95}}{2*2}=\frac{0+4\sqrt{95}}{4}
=\frac{4\sqrt{95}}{4}
=\sqrt{95}
$