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9x^2+39x+42=0
a = 9; b = 39; c = +42;
Δ = b2-4ac
Δ = 392-4·9·42
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-3}{2*9}=\frac{-42}{18} =-2+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+3}{2*9}=\frac{-36}{18} =-2 $
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