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4j^2+31j-8=0
a = 4; b = 31; c = -8;
Δ = b2-4ac
Δ = 312-4·4·(-8)
Δ = 1089
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{1089}=33$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-33}{2*4}=\frac{-64}{8} =-8 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+33}{2*4}=\frac{2}{8} =1/4 $
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