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10j^2+33j=0
a = 10; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·10·0
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$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*10}=\frac{-66}{20} =-3+3/10 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*10}=\frac{0}{20} =0 $
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