r^2+9r+13=0

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Solution for r^2+9r+13=0 equation: 



r^2+9r+13=0

a = 1; b = 9; c = +13;
Δ = b2-4ac
Δ = 92-4·1·13
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{29}}{2*1}=\frac{-9-\sqrt{29}}{2}
$


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

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