one-one
Let ${\mathit{x}}_1,{\mathit{x}}_2\in \mathit{R}+$ and $\mathit{f}\left({\mathit{x}}_1\right)=\mathit{f}\left({\mathit{x}}_2\right)$
${9{\mathit{x}}_1}^2+6{\mathit{x}}_1-5={9{\mathit{x}}_2}^2+6{\mathit{x}}_2-5$
${9{\mathit{x}}_1}^2+6{\mathit{x}}_1={9{\mathit{x}}_2}^2+6{\mathit{x}}_2$ (removing -5 from both side)
${3{\mathit{x}}_1}^2+2{\mathit{x}}_1={3{\mathit{x}}_2}^2+2{\mathit{x}}_2$ (dividing 3 on both side)
${3{\mathit{x}}_1}^2-{3{\mathit{x}}_2}^2+2{\mathit{x}}_1-2{\mathit{x}}_2=0$
$\Rightarrow 3\left({{\mathit{x}}_1}^2-{{\mathit{x}}_2}^2\right)+2\left({\mathit{x}}_1-{\mathit{x}}_2\right)=0$
$3\left({\mathit{x}}_1-{\mathit{x}}_2\right)\left({\mathit{x}}_1+{\mathit{x}}_2\right)+2\left({\mathit{x}}_1-{\mathit{x}}_2\right)=0$
${\mathit{x}}_1-{\mathit{x}}_2=0$ (OR) ${\mathit{x}}_1+{\mathit{x}}_2+2=0$
Consider ${\mathit{x}}_1+{\mathit{x}}_2+2=0\Rightarrow {\mathit{x}}_1=-{\mathit{x}}_2-2$
As ${\mathit{x}}_1,{\mathit{x}}_2\in \mathit{R}+$ (non-negative real numbers), it $\left({\mathit{x}}_1+{\mathit{x}}_2+2=0\right)$ can not be possible. So ${\mathit{x}}_1-{\mathit{x}}_2=0\Rightarrow {\mathit{x}}_1={\mathit{x}}_2$
Hence it is one-one.
onto
Let $\mathit{y}=\mathit{f}\left(\mathit{x}\right)$ and $\mathit{y}\in \left[-5,\mathit{\infty }\right]$
So $\mathit{y}=9{\mathit{x}}^2+6\mathit{x}-5$
$9{\mathit{x}}^2+6\mathit{x}-5-\mathit{y}=0$
$\mathit{x}=\frac{-6\pm \sqrt{6^2-4\times 9\times \left(-\mathit{y}-5\right)}}{2\times 9}$ (by using formula $\mathit{x}=\frac{-\mathit{b}\pm \sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}}{2\mathit{a}}$ for quadratic equation $\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}=0$)
$\mathit{x}=\frac{-6\pm \sqrt{36+36\left(\mathit{y}+5\right)}}{18}$
$\Rightarrow \mathit{x}=\frac{-6\pm 6\sqrt{\left(\mathit{y}+6\right)}}{18}$
x has two solution $\mathit{x}=\frac{-1-\sqrt{\mathit{y}+6}}{3}$ and $\mathit{x}=\frac{-1+\sqrt{\mathit{y}+6}}{3}$. We need to prove either one solution (x) exists for every $\mathit{y}\in \left[-5,\mathit{\infty }\right)$. In the solution $\mathit{x}=\frac{-1-\sqrt{\mathit{y}+6}}{3}$, x is the negative number for any y in the given range. Now we left with $\mathit{x}=\frac{-1+\sqrt{\mathit{y}+6}}{3}$. 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 $\mathit{y}\in \left[-5,\mathit{\infty }\right)$ there exists a value in domain $\to \mathit{x}\in \mathit{R}+$. Hence it is onto.
So the function is bijective (one-one and onto).