-r^2+10r-24=0

Solution for -r^2+10r-24=0 equation:

-r^2+10r-24=0

We add all the numbers together, and all the variables

-1r^2+10r-24=0

a = -1; b = 10; c = -24;
Δ = b2-4ac
Δ = 102-4·(-1)·(-24)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*-1}=\frac{-12}{-2}
=+6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*-1}=\frac{-8}{-2}
=+4
$