4m2-48=0

Solution for 4m2-48=0 equation:

4m^2-48=0

a = 4; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·4·(-48)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*4}=\frac{0-16\sqrt{3}}{8}
=-\frac{16\sqrt{3}}{8}
=-2\sqrt{3}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*4}=\frac{0+16\sqrt{3}}{8}
=\frac{16\sqrt{3}}{8}
=2\sqrt{3}
$