2d^2+15d+20=0

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Solution for 2d^2+15d+20=0 equation: 



2d^2+15d+20=0

a = 2; b = 15; c = +20;
Δ = b2-4ac
Δ = 152-4·2·20
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{65}}{2*2}=\frac{-15-\sqrt{65}}{4}
$


$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{65}}{2*2}=\frac{-15+\sqrt{65}}{4}
$

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