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8k^2+29k+16=0
a = 8; b = 29; c = +16;
Δ = b2-4ac
Δ = 292-4·8·16
Δ = 329
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{-(29)-\sqrt{329}}{2*8}=\frac{-29-\sqrt{329}}{16} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{329}}{2*8}=\frac{-29+\sqrt{329}}{16} $
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