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27x^2+5x-9=0
a = 27; b = 5; c = -9;
Δ = b2-4ac
Δ = 52-4·27·(-9)
Δ = 997
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{-(5)-\sqrt{997}}{2*27}=\frac{-5-\sqrt{997}}{54} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{997}}{2*27}=\frac{-5+\sqrt{997}}{54} $
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