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-200=30j-5)5j+20
We move all terms to the left:
-200-(30j-5)5j+20)=0
We add all the numbers together, and all the variables
-(30j-5)5j=0
We multiply parentheses
-150j^2+25j=0
a = -150; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·(-150)·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$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*-150}=\frac{-50}{-300} =1/6 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*-150}=\frac{0}{-300} =0 $
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