3r2+8r+5=0

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Solution for 3r2+8r+5=0 equation: 



3r^2+8r+5=0

a = 3; b = 8; c = +5;
Δ = b2-4ac
Δ = 82-4·3·5
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


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


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

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