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9x^2-42x+40=0
a = 9; b = -42; c = +40;
Δ = b2-4ac
Δ = -422-4·9·40
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-18}{2*9}=\frac{24}{18} =1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+18}{2*9}=\frac{60}{18} =3+1/3 $
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