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j2+2j-24=0
We add all the numbers together, and all the variables
j^2+2j-24=0
a = 1; b = 2; c = -24;
Δ = b2-4ac
Δ = 22-4·1·(-24)
Δ = 100
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{100}=10$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*1}=\frac{-12}{2} =-6 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*1}=\frac{8}{2} =4 $
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