12m^2+19m-21=0

Solution for 12m^2+19m-21=0 equation:

12m^2+19m-21=0

a = 12; b = 19; c = -21;
Δ = b2-4ac
Δ = 192-4·12·(-21)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-37}{2*12}=\frac{-56}{24}
=-2+1/3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+37}{2*12}=\frac{18}{24}
=3/4
$