4m^2+25m-56=0

Solution for 4m^2+25m-56=0 equation:

4m^2+25m-56=0

a = 4; b = 25; c = -56;
Δ = b2-4ac
Δ = 252-4·4·(-56)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-39}{2*4}=\frac{-64}{8}
=-8
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+39}{2*4}=\frac{14}{8}
=1+3/4
$