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32j^2+43j=0
a = 32; b = 43; c = 0;
Δ = b2-4ac
Δ = 432-4·32·0
Δ = 1849
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{1849}=43$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-43}{2*32}=\frac{-86}{64} =-1+11/32 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+43}{2*32}=\frac{0}{64} =0 $
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