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13j^2+32j=0
a = 13; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·13·0
Δ = 1024
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{1024}=32$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*13}=\frac{-64}{26} =-2+6/13 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*13}=\frac{0}{26} =0 $
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