12z^2+19z=-4

Solution for 12z^2+19z=-4 equation:

12z^2+19z=-4

We move all terms to the left:

12z^2+19z-(-4)=0

We add all the numbers together, and all the variables

12z^2+19z+4=0

a = 12; b = 19; c = +4;
Δ = b2-4ac
Δ = 192-4·12·4
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*12}=\frac{-32}{24}
=-1+1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*12}=\frac{-6}{24}
=-1/4
$