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2500=x^2+x
We move all terms to the left:
2500-(x^2+x)=0
We get rid of parentheses
-x^2-x+2500=0
We add all the numbers together, and all the variables
-1x^2-1x+2500=0
a = -1; b = -1; c = +2500;
Δ = b2-4ac
Δ = -12-4·(-1)·2500
Δ = 10001
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{10001}}{2*-1}=\frac{1-\sqrt{10001}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{10001}}{2*-1}=\frac{1+\sqrt{10001}}{-2} $
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