-r^2+4r+45=0

Solution for -r^2+4r+45=0 equation:

-r^2+4r+45=0

We add all the numbers together, and all the variables

-1r^2+4r+45=0

a = -1; b = 4; c = +45;
Δ = b2-4ac
Δ = 42-4·(-1)·45
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{-(4)-14}{2*-1}=\frac{-18}{-2}
=+9
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*-1}=\frac{10}{-2}
=-5
$