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6k^2+36k+48=0
a = 6; b = 36; c = +48;
Δ = b2-4ac
Δ = 362-4·6·48
Δ = 144
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{144}=12$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-12}{2*6}=\frac{-48}{12} =-4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+12}{2*6}=\frac{-24}{12} =-2 $
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