r^2+18r-180=0

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Solution for r^2+18r-180=0 equation:



r^2+18r-180=0
a = 1; b = 18; c = -180;
Δ = b2-4ac
Δ = 182-4·1·(-180)
Δ = 1044
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:
$\sqrt{\Delta}=\sqrt{1044}=\sqrt{36*29}=\sqrt{36}*\sqrt{29}=6\sqrt{29}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{29}}{2*1}=\frac{-18-6\sqrt{29}}{2} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{29}}{2*1}=\frac{-18+6\sqrt{29}}{2} $

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