3r2-2r-5=0

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Solution for 3r2-2r-5=0 equation: 



3r^2-2r-5=0

a = 3; b = -2; c = -5;
Δ = b2-4ac
Δ = -22-4·3·(-5)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*3}=\frac{-6}{6}
=-1
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*3}=\frac{10}{6}
=1+2/3
$

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