2r^2+7r+3=0

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Solution for 2r^2+7r+3=0 equation: 



2r^2+7r+3=0

a = 2; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·2·3
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5}{2*2}=\frac{-12}{4}
=-3
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*2}=\frac{-2}{4}
=-1/2
$

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