15k^2+19k-6=0

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Solution for 15k^2+19k-6=0 equation: 



15k^2+19k-6=0

a = 15; b = 19; c = -6;
Δ = b2-4ac
Δ = 192-4·15·(-6)
Δ = 721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{721}}{2*15}=\frac{-19-\sqrt{721}}{30}
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{721}}{2*15}=\frac{-19+\sqrt{721}}{30}
$

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