3r^2+r-5=0

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



3r^2+r-5=0

a = 3; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·3·(-5)
Δ = 61
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{61}}{2*3}=\frac{-1-\sqrt{61}}{6}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{61}}{2*3}=\frac{-1+\sqrt{61}}{6}
$

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