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12k^2-18=7k
We move all terms to the left:
12k^2-18-(7k)=0
a = 12; b = -7; c = -18;
Δ = b2-4ac
Δ = -72-4·12·(-18)
Δ = 913
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{913}}{2*12}=\frac{7-\sqrt{913}}{24} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{913}}{2*12}=\frac{7+\sqrt{913}}{24} $
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