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x+100x^2=2500
We move all terms to the left:
x+100x^2-(2500)=0
a = 100; b = 1; c = -2500;
Δ = b2-4ac
Δ = 12-4·100·(-2500)
Δ = 1000001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1000001}}{2*100}=\frac{-1-\sqrt{1000001}}{200} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1000001}}{2*100}=\frac{-1+\sqrt{1000001}}{200} $
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