r^2-13r+37=0

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Solution for r^2-13r+37=0 equation: 



r^2-13r+37=0

a = 1; b = -13; c = +37;
Δ = b2-4ac
Δ = -132-4·1·37
Δ = 21
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{-(-13)-\sqrt{21}}{2*1}=\frac{13-\sqrt{21}}{2}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{21}}{2*1}=\frac{13+\sqrt{21}}{2}
$

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