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6k^2-21k+18=0
a = 6; b = -21; c = +18;
Δ = b2-4ac
Δ = -212-4·6·18
Δ = 9
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{9}=3$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3}{2*6}=\frac{18}{12} =1+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3}{2*6}=\frac{24}{12} =2 $
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