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6k^2+20k-50=0
a = 6; b = 20; c = -50;
Δ = b2-4ac
Δ = 202-4·6·(-50)
Δ = 1600
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{1600}=40$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-40}{2*6}=\frac{-60}{12} =-5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+40}{2*6}=\frac{20}{12} =1+2/3 $
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