3m^2+23m-8=0

Solution for 3m^2+23m-8=0 equation:

3m^2+23m-8=0

a = 3; b = 23; c = -8;
Δ = b2-4ac
Δ = 232-4·3·(-8)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*3}=\frac{-48}{6}
=-8
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*3}=\frac{2}{6}
=1/3
$