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9x^2+41x+10=0
a = 9; b = 41; c = +10;
Δ = b2-4ac
Δ = 412-4·9·10
Δ = 1321
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{-(41)-\sqrt{1321}}{2*9}=\frac{-41-\sqrt{1321}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{1321}}{2*9}=\frac{-41+\sqrt{1321}}{18} $
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