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9x^2-490x+2000=0
a = 9; b = -490; c = +2000;
Δ = b2-4ac
Δ = -4902-4·9·2000
Δ = 168100
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{168100}=410$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-490)-410}{2*9}=\frac{80}{18} =4+4/9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-490)+410}{2*9}=\frac{900}{18} =50 $
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