# How do you solve 7v^2+1=29?

$v = \setminus \pm 2$

#### Explanation:

Given that

$7 {v}^{2} + 1 = 29$

$7 {v}^{2} = 29 - 1$

$7 {v}^{2} = 28$

${v}^{2} = \frac{28}{7}$

${v}^{2} = 4$

$\setminus \sqrt{{v}^{2}} = \setminus \sqrt{4} \setminus \quad \left(\setminus \textrm{t a k \in g \square \sqrt[o]{n} \bot h s i \mathrm{de} s}\right)$

$| v | = 2$

$v = \setminus \pm 2$

Jul 17, 2018

$v = \pm 2$

#### Explanation:

$\text{subtract 1 from both sides}$

$7 {v}^{2} = 29 - 1 = 28$

$\text{divide both sides by 7}$

${v}^{2} = \frac{28}{7} = 4$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$v = \pm \sqrt{4} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow v = \pm 2$

Jul 17, 2018

$v = \pm 2$

#### Explanation:

Remember, our end goal is to isolate $v$. We can start by subtracting $1$ from both sides. We now have

$7 {v}^{2} = 28$

Dividing both sides by $7$ gives us

${v}^{2} = 4$

Taking the $\pm$ square root of both sides, we get

$v = \pm 2$

Hope this helps!