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6k^2+4k-14=0
a = 6; b = 4; c = -14;
Δ = b2-4ac
Δ = 42-4·6·(-14)
Δ = 352
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{22}}{2*6}=\frac{-4-4\sqrt{22}}{12} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{22}}{2*6}=\frac{-4+4\sqrt{22}}{12} $
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