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j^2+14j+13=0
a = 1; b = 14; c = +13;
Δ = b2-4ac
Δ = 142-4·1·13
Δ = 144
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{144}=12$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-12}{2*1}=\frac{-26}{2} =-13 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+12}{2*1}=\frac{-2}{2} =-1 $
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