3r^2+5r-8=0

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Solution for 3r^2+5r-8=0 equation: 



3r^2+5r-8=0

a = 3; b = 5; c = -8;
Δ = b2-4ac
Δ = 52-4·3·(-8)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*3}=\frac{-16}{6}
=-2+2/3
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*3}=\frac{6}{6}
=1
$

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