5m^2+19m=-12

Solution for 5m^2+19m=-12 equation:

5m^2+19m=-12

We move all terms to the left:

5m^2+19m-(-12)=0

We add all the numbers together, and all the variables

5m^2+19m+12=0

a = 5; b = 19; c = +12;
Δ = b2-4ac
Δ = 192-4·5·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*5}=\frac{-30}{10}
=-3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*5}=\frac{-8}{10}
=-4/5
$