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2k^2+26k+24=0
a = 2; b = 26; c = +24;
Δ = b2-4ac
Δ = 262-4·2·24
Δ = 484
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{484}=22$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-22}{2*2}=\frac{-48}{4} =-12 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+22}{2*2}=\frac{-4}{4} =-1 $
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