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24k^2+8k-16=0
a = 24; b = 8; c = -16;
Δ = b2-4ac
Δ = 82-4·24·(-16)
Δ = 1600
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{1600}=40$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-40}{2*24}=\frac{-48}{48} =-1 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+40}{2*24}=\frac{32}{48} =2/3 $
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