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4j^2+8j-5=0
a = 4; b = 8; c = -5;
Δ = b2-4ac
Δ = 82-4·4·(-5)
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$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*4}=\frac{-20}{8} =-2+1/2 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*4}=\frac{4}{8} =1/2 $
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