-r^2+3r+4=0

Solution for -r^2+3r+4=0 equation:

-r^2+3r+4=0

We add all the numbers together, and all the variables

-1r^2+3r+4=0

a = -1; b = 3; c = +4;
Δ = b2-4ac
Δ = 32-4·(-1)·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*-1}=\frac{-8}{-2}
=+4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*-1}=\frac{2}{-2}
=-1
$