2k2+15k+7=0

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Solution for 2k2+15k+7=0 equation:



2k^2+15k+7=0
a = 2; b = 15; c = +7;
Δ = b2-4ac
Δ = 152-4·2·7
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-13}{2*2}=\frac{-28}{4} =-7 $
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+13}{2*2}=\frac{-2}{4} =-1/2 $

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