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9x^2+3x-11=0
a = 9; b = 3; c = -11;
Δ = b2-4ac
Δ = 32-4·9·(-11)
Δ = 405
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{405}=\sqrt{81*5}=\sqrt{81}*\sqrt{5}=9\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9\sqrt{5}}{2*9}=\frac{-3-9\sqrt{5}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9\sqrt{5}}{2*9}=\frac{-3+9\sqrt{5}}{18} $
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