-4m^2+8+32=0

Solution for -4m^2+8+32=0 equation:

-4m^2+8+32=0

We add all the numbers together, and all the variables

-4m^2+40=0

a = -4; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-4)·40
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{10}}{2*-4}=\frac{0-8\sqrt{10}}{-8}
=-\frac{8\sqrt{10}}{-8}
=-\frac{\sqrt{10}}{-1}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{10}}{2*-4}=\frac{0+8\sqrt{10}}{-8}
=\frac{8\sqrt{10}}{-8}
=\frac{\sqrt{10}}{-1}
$