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9x^2-43x+40=0
a = 9; b = -43; c = +40;
Δ = b2-4ac
Δ = -432-4·9·40
Δ = 409
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{-(-43)-\sqrt{409}}{2*9}=\frac{43-\sqrt{409}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+\sqrt{409}}{2*9}=\frac{43+\sqrt{409}}{18} $
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