42+8=2m^2

Solution for 42+8=2m^2 equation:

42+8=2m^2

We move all terms to the left:

42+8-(2m^2)=0

We add all the numbers together, and all the variables

-2m^2+50=0

a = -2; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-2)·50
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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