r^2-5r-3=0

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Solution for r^2-5r-3=0 equation: 



r^2-5r-3=0

a = 1; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·1·(-3)
Δ = 37
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{-(-5)-\sqrt{37}}{2*1}=\frac{5-\sqrt{37}}{2}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{37}}{2*1}=\frac{5+\sqrt{37}}{2}
$

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