-3r^2+22r-24=0

Solution for -3r^2+22r-24=0 equation:

-3r^2+22r-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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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