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=50V-5V^2
We move all terms to the left:
-(50V-5V^2)=0
We get rid of parentheses
5V^2-50V=0
a = 5; b = -50; c = 0;
Δ = b2-4ac
Δ = -502-4·5·0
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2500}=50$$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-50}{2*5}=\frac{0}{10} =0 $$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+50}{2*5}=\frac{100}{10} =10 $
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