r^2+11r-26=0

Solution for r^2+11r-26=0 equation:

r^2+11r-26=0

a = 1; b = 11; c = -26;
Δ = b2-4ac
Δ = 112-4·1·(-26)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-15}{2*1}=\frac{-26}{2}
=-13
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+15}{2*1}=\frac{4}{2}
=2
$