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j^2-14j-15=0
a = 1; b = -14; c = -15;
Δ = b2-4ac
Δ = -142-4·1·(-15)
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$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-16}{2*1}=\frac{-2}{2} =-1 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+16}{2*1}=\frac{30}{2} =15 $
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