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j^2-18j+17=0
a = 1; b = -18; c = +17;
Δ = b2-4ac
Δ = -182-4·1·17
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-16}{2*1}=\frac{2}{2} =1 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+16}{2*1}=\frac{34}{2} =17 $
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