r2-5r-36=0

Solution for r2-5r-36=0 equation:

r2-5r-36=0

We add all the numbers together, and all the variables

r^2-5r-36=0

a = 1; b = -5; c = -36;
Δ = b2-4ac
Δ = -52-4·1·(-36)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*1}=\frac{-8}{2}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*1}=\frac{18}{2}
=9
$