-r^2-3r+54=0

Solution for -r^2-3r+54=0 equation:

-r^2-3r+54=0

We add all the numbers together, and all the variables

-1r^2-3r+54=0

a = -1; b = -3; c = +54;
Δ = b2-4ac
Δ = -32-4·(-1)·54
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*-1}=\frac{-12}{-2}
=+6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*-1}=\frac{18}{-2}
=-9
$