412=4m2+12

Solution for 412=4m2+12 equation:

412=4m^2+12

We move all terms to the left:

412-(4m^2+12)=0

We get rid of parentheses

-4m^2-12+412=0

We add all the numbers together, and all the variables

-4m^2+400=0

a = -4; b = 0; c = +400;
Δ = b2-4ac
Δ = 02-4·(-4)·400
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*-4}=\frac{-80}{-8}
=+10
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*-4}=\frac{80}{-8}
=-10
$