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(8j^2)-21j=0
a = 8; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·8·0
Δ = 441
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{441}=21$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*8}=\frac{0}{16} =0 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*8}=\frac{42}{16} =2+5/8 $
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