![a + 8 = - \frac 5 a](https://img.qammunity.org/2019/formulas/mathematics/high-school/mbzoe2sg556qcow0cxsb4xjn1zr47x7pj0.png)
If I was solving this, which I suppose I am, my first step would be the clear the fractions by multiplying both sides by a.
![a^2 + 8a = - 5](https://img.qammunity.org/2019/formulas/mathematics/high-school/p6jqybqn4fii5h9g1b3gyvec0ntibrrde2.png)
That's there, first choice, SELECT IT
Then I'd bring the 5 to the other side;
![a^2 + 8a + 5 = 0](https://img.qammunity.org/2019/formulas/mathematics/high-school/jzp87jsibw1jaowey8l0r0hjft8546xwlb.png)
Not among the choices.
Then I'd attempt to factor, looking for a pair that multiplies to 5 and adds to 8; no joy there.
Shakespeare Quadratic Formula (2b or -2b) trick:
![x^2 - 2bx + c \textrm{ has zeros at } x=b\pm √(b^2-c)](https://img.qammunity.org/2019/formulas/mathematics/high-school/5qq2igcqh46s20pu3v4e649m2ql214w112.png)
![a = 4 \pm √(4^2-5) = 4\pm √(11)](https://img.qammunity.org/2019/formulas/mathematics/high-school/krvzu6jvxzqhgrlf8v82zv4ksh6tx2eupt.png)
All the other choices have nice rational answers so can't be right.
First choice only