5r^2+8r+3=0

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



5r^2+8r+3=0

a = 5; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·5·3
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*5}=\frac{-10}{10}
=-1
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*5}=\frac{-6}{10}
=-3/5
$

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