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(9x^2+19x+2)=0
We get rid of parentheses
9x^2+19x+2=0
a = 9; b = 19; c = +2;
Δ = b2-4ac
Δ = 192-4·9·2
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*9}=\frac{-36}{18} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*9}=\frac{-2}{18} =-1/9 $
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