-r^2+5r+6=0

Solution for -r^2+5r+6=0 equation:

-r^2+5r+6=0

We add all the numbers together, and all the variables

-1r^2+5r+6=0

a = -1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·(-1)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-1}=\frac{-12}{-2}
=+6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-1}=\frac{2}{-2}
=-1
$