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9x^2+41x-204=0
a = 9; b = 41; c = -204;
Δ = b2-4ac
Δ = 412-4·9·(-204)
Δ = 9025
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{9025}=95$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-95}{2*9}=\frac{-136}{18} =-7+5/9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+95}{2*9}=\frac{54}{18} =3 $
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