r^2+20r-261=0

Solution for r^2+20r-261=0 equation:

r^2+20r-261=0

a = 1; b = 20; c = -261;
Δ = b2-4ac
Δ = 202-4·1·(-261)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-38}{2*1}=\frac{-58}{2}
=-29
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+38}{2*1}=\frac{18}{2}
=9
$