# How do you evaluate sqrt(b^2-4ac) for a=2, b=-5, c=2?

Feb 28, 2017

$\pm 3$

#### Explanation:

$\sqrt{{b}^{2} - 4 a c} \text{ "->" } \sqrt{{\left(- 5\right)}^{2} - 4 \left(2\right) \left(2\right)}$

$\sqrt{+ 25 - 16} = \sqrt{9} = \pm 3$

Feb 28, 2017

See the entire solution process below:

#### Explanation:

To solve this you must substitute:

$\textcolor{red}{2}$ for $\textcolor{red}{a}$ in the expression.

$\textcolor{b l u e}{- 5}$ for $\textcolor{b l u e}{b}$ in the expression.

$\textcolor{p u r p \le}{2}$ for $\textcolor{p u r p \le}{c}$ in the expression.

$\sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{p u r p \le}{c}}$ becomes:

$\sqrt{{\textcolor{b l u e}{- 5}}^{2} - \left(4 \times \textcolor{red}{2} \times \textcolor{p u r p \le}{2}\right)}$

$\sqrt{25 - 16}$

$\sqrt{9} = \pm 3$

• Reminder, the square root of a number results in plus or minus the result.