one-oneLet $x_1, x_2 \in \mathbb{R}_+$ and $f(x_1) = f(x_2)$$(9x_1)^2 + 6x_1 -5 = (9x_2)^2 + 6x_2 - 5$$(9x_1)^2 + 6x_1 = (9x_2)^2 + 6x_2$ (removing -5 from both side)$(3x_1)^2 + 2x_1 = (3x_2)^2 + 2x_2$ (dividing 3 on both side)$(3x_1)^2 - (3x_2)^2 + 2x_1 - 2x_2 = 0$$\Rightarrow 3(x_1^2 - x_2^2) + 2(x_1 - x_2) = 0$$3(x_1 - x_2)(x_1 + x_2) + 2(x_1 - x_2) = 0$$x_1 - x_2 = 0$ (OR) $x_1 + x_2 + 2 = 0$Consider $x_1 + x_2 + 2 = 0 \Rightarrow x_1 = -x_2 - 2$As $x_1, x_2 \in \mathbb{R}_+$ (non-negative real numbers), it $(x_1 + x_2 + 2 = 0)$ can not be possible. So $x_1 - x_2 = 0 \Rightarrow x_1 = x_2$Hence it is one-one.ontoLet $y = f(x)$ and $y \in [-5, \infty]$So $y = 9x^2 + 6x - 5$$9x^2 + 6x - 5 - y = 0$$x = \frac{-6 \pm \sqrt{6^2 - 4 \times 9 \times (-y -5)}}{2 \times 9}$ (by using formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for quadratic equation $ax^2 + bx + c = 0$)$x = \frac{-6 \pm \sqrt{36 + 36(y+5)}}{18}$$\Rightarrow x = \frac{-6 \pm 6\sqrt{y+6}}{18}$x has two solution $x = \frac{-1 - \sqrt{y+6}}{3}$ and $x = \frac{-1 + \sqrt{y+6}}{3}$. We need to prove either one solution (x) exists for every $y \in [-5, \infty)$. In the solution $x = \frac{-1 - \sqrt{y+6}}{3}$, x is the negative number for any y in the given range. Now we left with $x = \frac{-1 + \sqrt{y+6}}{3}$. For y = -5, x = 0 and any value greater then -5 x is always non negative real numbers. It proves that every value in the range $y \in [-5, \infty)$ there exists a value in domain $\rightarrow x \in \mathbb{R}_+$. Hence it is onto.So the function is bijective (one-one and onto).