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j2=2.5
We move all terms to the left:
j2-(2.5)=0
We add all the numbers together, and all the variables
j^2-2.5=0
a = 1; b = 0; c = -2.5;
Δ = b2-4ac
Δ = 02-4·1·(-2.5)
Δ = 10
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}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{10}}{2*1}=\frac{0-\sqrt{10}}{2} =-\frac{\sqrt{}}{2} $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{10}}{2*1}=\frac{0+\sqrt{10}}{2} =\frac{\sqrt{}}{2} $
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