412=4m^2=+12

Solution for 412=4m^2=+12 equation:

412=4m^2=+12

We move all terms to the left:

412-(4m^2)=0

a = -4; b = 0; c = +412;
Δ = b2-4ac
Δ = 02-4·(-4)·412
Δ = 6592
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:

$\sqrt{\Delta}=\sqrt{6592}=\sqrt{64*103}=\sqrt{64}*\sqrt{103}=8\sqrt{103}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{103}}{2*-4}=\frac{0-8\sqrt{103}}{-8}
=-\frac{8\sqrt{103}}{-8}
=-\frac{\sqrt{103}}{-1}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{103}}{2*-4}=\frac{0+8\sqrt{103}}{-8}
=\frac{8\sqrt{103}}{-8}
=\frac{\sqrt{103}}{-1}
$