r^2+17r+72=0

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Solution for r^2+17r+72=0 equation: 



r^2+17r+72=0

a = 1; b = 17; c = +72;
Δ = b2-4ac
Δ = 172-4·1·72
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-1}{2*1}=\frac{-18}{2}
=-9
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*1}=\frac{-16}{2}
=-8
$

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