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9x^2-28x+11=0
a = 9; b = -28; c = +11;
Δ = b2-4ac
Δ = -282-4·9·11
Δ = 388
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{97}}{2*9}=\frac{28-2\sqrt{97}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{97}}{2*9}=\frac{28+2\sqrt{97}}{18} $
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