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6k^2-34k+48=0
a = 6; b = -34; c = +48;
Δ = b2-4ac
Δ = -342-4·6·48
Δ = 4
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{4}=2$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-2}{2*6}=\frac{32}{12} =2+2/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+2}{2*6}=\frac{36}{12} =3 $
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