r^2+5r-23=0

Solution for r^2+5r-23=0 equation:

r^2+5r-23=0

a = 1; b = 5; c = -23;
Δ = b2-4ac
Δ = 52-4·1·(-23)
Δ = 117
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{13}}{2*1}=\frac{-5-3\sqrt{13}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{13}}{2*1}=\frac{-5+3\sqrt{13}}{2}
$