r^2+r-3=0

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



r^2+r-3=0

a = 1; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·1·(-3)
Δ = 13
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{-(1)-\sqrt{13}}{2*1}=\frac{-1-\sqrt{13}}{2}
$


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

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