-3z^2+22z=24

Solution for -3z^2+22z=24 equation:

-3z^2+22z=24

We move all terms to the left:

-3z^2+22z-(24)=0

a = -3; b = 22; c = -24;
Δ = b2-4ac
Δ = 222-4·(-3)·(-24)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*-3}=\frac{-36}{-6}
=+6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*-3}=\frac{-8}{-6}
=1+1/3
$