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k^2-16k+15=0
a = 1; b = -16; c = +15;
Δ = b2-4ac
Δ = -162-4·1·15
Δ = 196
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{196}=14$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-14}{2*1}=\frac{2}{2} =1 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+14}{2*1}=\frac{30}{2} =15 $
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