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5k^2-3k-5=0
a = 5; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·5·(-5)
Δ = 109
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{-(-3)-\sqrt{109}}{2*5}=\frac{3-\sqrt{109}}{10} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{109}}{2*5}=\frac{3+\sqrt{109}}{10} $
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