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x^2-2x+20=140
We move all terms to the left:
x^2-2x+20-(140)=0
We add all the numbers together, and all the variables
x^2-2x-120=0
a = 1; b = -2; c = -120;
Δ = b2-4ac
Δ = -22-4·1·(-120)
Δ = 484
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*1}=\frac{-20}{2} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*1}=\frac{24}{2} =12 $
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