2d^2+17d+15=0

Solution for 2d^2+17d+15=0 equation:

2d^2+17d+15=0

a = 2; b = 17; c = +15;
Δ = b2-4ac
Δ = 172-4·2·15
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-13}{2*2}=\frac{-30}{4}
=-7+1/2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*2}=\frac{-4}{4}
=-1
$