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5k^2+26k-24=0
a = 5; b = 26; c = -24;
Δ = b2-4ac
Δ = 262-4·5·(-24)
Δ = 1156
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{1156}=34$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-34}{2*5}=\frac{-60}{10} =-6 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+34}{2*5}=\frac{8}{10} =4/5 $
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