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20x^2=x+40
We move all terms to the left:
20x^2-(x+40)=0
We get rid of parentheses
20x^2-x-40=0
We add all the numbers together, and all the variables
20x^2-1x-40=0
a = 20; b = -1; c = -40;
Δ = b2-4ac
Δ = -12-4·20·(-40)
Δ = 3201
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{3201}}{2*20}=\frac{1-\sqrt{3201}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{3201}}{2*20}=\frac{1+\sqrt{3201}}{40} $
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