19z+2z^2+24=0

Solution for 19z+2z^2+24=0 equation:

19z+2z^2+24=0

a = 2; b = 19; c = +24;
Δ = b2-4ac
Δ = 192-4·2·24
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*2}=\frac{-32}{4}
=-8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*2}=\frac{-6}{4}
=-1+1/2
$