4m^2+48m=-128

Solution for 4m^2+48m=-128 equation:

4m^2+48m=-128

We move all terms to the left:

4m^2+48m-(-128)=0

We add all the numbers together, and all the variables

4m^2+48m+128=0

a = 4; b = 48; c = +128;
Δ = b2-4ac
Δ = 482-4·4·128
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16}{2*4}=\frac{-64}{8}
=-8
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16}{2*4}=\frac{-32}{8}
=-4
$