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2k^2-14=3k
We move all terms to the left:
2k^2-14-(3k)=0
a = 2; b = -3; c = -14;
Δ = b2-4ac
Δ = -32-4·2·(-14)
Δ = 121
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{121}=11$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*2}=\frac{-8}{4} =-2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*2}=\frac{14}{4} =3+1/2 $
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