18m^2+48m-256=0

Solution for 18m^2+48m-256=0 equation:

18m^2+48m-256=0

a = 18; b = 48; c = -256;
Δ = b2-4ac
Δ = 482-4·18·(-256)
Δ = 20736
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{20736}=144$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-144}{2*18}=\frac{-192}{36}
=-5+1/3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+144}{2*18}=\frac{96}{36}
=2+2/3
$