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

Nov 30, 2017

In fraction form:

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

Simplified form:

$\setminus \rightarrow x \setminus \approx - 1.75 \setminus \pm 1.19895788 i$

#### Explanation:

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

$2 {x}^{2} + 7 x + 9 = 0$

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

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

Now, move $c$ to the RHS:

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

Now add ${\left(\setminus \frac{b}{2}\right)}^{2}$ to both sides. Here, $b = 7$.

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

Simplify the RHS:

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

Factor the LHS into ${\left(x + \setminus \frac{b}{2}\right)}^{2}$:

$\setminus \rightarrow {\left(x + \setminus \frac{\setminus \frac{7}{2}}{2}\right)}^{2} = - \setminus \frac{23}{16}$

$\setminus \rightarrow {\left(x + \setminus \frac{7}{4}\right)}^{2} = - \setminus \frac{23}{16}$

Take the square root of both sides:

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

Isolate $x$:

$\setminus \rightarrow x = \setminus \sqrt{- \setminus \frac{23}{16}} - \setminus \frac{7}{4}$

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

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

$\setminus \rightarrow x \setminus \approx - 1.75 \setminus \pm 1.19895788 i$

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

$2 {x}^{2} + 7 x + 9 = 0$

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

$a = 2$

$b = 7$

$c = 9$

Now, plug them into the formula:

$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\setminus \rightarrow x = \setminus \frac{- 7 \setminus \pm \setminus \sqrt{{7}^{2} - 4 \left(2\right) \left(9\right)}}{2 \left(2\right)}$

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

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

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

Rearranging yields:

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

$\setminus \rightarrow x \setminus \approx - 1.75 \setminus \pm 1.19895788 i$

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