# How do you evaluate (x-y)^2-2wz if w=6, x=0.4, y=1/2, z=-3?

Apr 29, 2017

See the solution process below:

#### Explanation:

We need to do the following substitutions into the expression:

$\textcolor{red}{6}$ for $\textcolor{red}{w}$

$\textcolor{b l u e}{0.4}$ for $\textcolor{b l u e}{x}$

$\textcolor{g r e e n}{\frac{1}{2}}$ for $\textcolor{g r e e n}{y}$

$\textcolor{p u r p \le}{- 3}$ for $\textcolor{p u r p \le}{z}$

${\left(\textcolor{b l u e}{x} - \textcolor{g r e e n}{y}\right)}^{2} - 2 \textcolor{red}{w} \textcolor{p u r p \le}{z}$ becomes:

${\left(\textcolor{b l u e}{0.4} - \textcolor{g r e e n}{\frac{1}{2}}\right)}^{2} - \left(2 \cdot \textcolor{red}{6} \cdot \textcolor{p u r p \le}{- 3}\right) \implies$

${\left(\textcolor{b l u e}{0.4} - \textcolor{g r e e n}{0.5}\right)}^{2} - \left(12 \cdot \textcolor{p u r p \le}{- 3}\right) \implies$

${\left(- 0.1\right)}^{2} - \left(- 36\right) \implies$

$0.01 + 36 \implies$

$36.01$