How do you solve `2x^2+7x+9=0` algebraically?

We’ll solve this two ways: by completing the square and using the quadratic formula. Here’s the first way:

`2x^2+7x+9=0`

First divide everything by `2`, since `a` cannot have a coefficient:

`\rightarrow x^2+\frac{7}{2}x+\frac{9}{2}=0`

Now, move `c` to the RHS:

`\rightarrow x^2+\frac{7}{2}x=-\frac{9}{2}`

Now add `(\frac{b}{2})^2` to both sides. Here, `b=7`.

`\rightarrow x^2+\frac{7}{2}x+\frac{49}{16}=\frac{49}{16}-\frac{9}{2}`

Simplify the RHS:

`\rightarrow x^2+\frac{7}{2}x+\frac{49}{16}=-\frac{23}{16}`

Factor the LHS into `(x+\frac{b}{2})^2`:

`\rightarrow (x+\frac{\frac{7}{2}}{2})^2=-\frac{23}{16}`

`\rightarrow (x+\frac{7}{4})^2=-\frac{23}{16}`

Take the square root of both sides:

`\rightarrow x+\frac{7}{4}=\sqrt{-\frac{23}{16}}`

Isolate `x`:

`\rightarrow x=\sqrt{-\frac{23}{16}}-\frac{7}[4}`

`\rightarrow x=i\sqrt{\frac{23}{16}}-\frac{7}{4}`

`\rightarrow x=i\frac{\sqrt{23}}{4}-\frac{7}{4}`

`\rightarrow x\approx-1.75\pm 1.19895788i`

Here’s the second way, using the quadratic formula:

`2x^2+7x+9=0`

First, we need to identify `a`, `b`, and `c`:

`a=2`

`b=7`

`c=9`

Now, plug them into the formula:

`x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}`

`\rightarrow x=\frac{-7\pm\sqrt{7^2-4(2)(9)}}{2(2)}`

`\rightarrow x=\frac{-7\pm\sqrt{49-72}}{4}`

`\rightarrow x=\frac{-7\pm\sqrt{-23}}{4}`

`\rightarrow x=\frac{-7\pm i\sqrt{23}}{4}`

Rearranging yields:

`\rightarrow x=\frac{i\sqrt{23}}{4}-\frac{7}{4}`

`\rightarrow x\approx-1.75\pm 1.19895788i`

It’s the same answer we got by completing the square, so we know the answer is correct.