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k2+6k-16=0
We add all the numbers together, and all the variables
k^2+6k-16=0
a = 1; b = 6; c = -16;
Δ = b2-4ac
Δ = 62-4·1·(-16)
Δ = 100
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{100}=10$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*1}=\frac{-16}{2} =-8 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*1}=\frac{4}{2} =2 $
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