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4k^2+6k-18=0
a = 4; b = 6; c = -18;
Δ = b2-4ac
Δ = 62-4·4·(-18)
Δ = 324
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{324}=18$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*4}=\frac{-24}{8} =-3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*4}=\frac{12}{8} =1+1/2 $
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