2r^2+14r+20=0

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Solution for 2r^2+14r+20=0 equation: 



2r^2+14r+20=0

a = 2; b = 14; c = +20;
Δ = b2-4ac
Δ = 142-4·2·20
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6}{2*2}=\frac{-20}{4}
=-5
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6}{2*2}=\frac{-8}{4}
=-2
$

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