-3z^2+21z+24=0

Solution for -3z^2+21z+24=0 equation:

-3z^2+21z+24=0

a = -3; b = 21; c = +24;
Δ = b2-4ac
Δ = 212-4·(-3)·24
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-27}{2*-3}=\frac{-48}{-6}
=+8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+27}{2*-3}=\frac{6}{-6}
=-1
$