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32j^2+25j=0
a = 32; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·32·0
Δ = 625
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{625}=25$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*32}=\frac{-50}{64} =-25/32 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*32}=\frac{0}{64} =0 $
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