# How do you solve 0.4t^2+0.7t=0.3t-0.2 by completing the square?

Nov 27, 2017

$t = - 0.5 \pm 0.5 i$

#### Explanation:

$0.4 {t}^{2} + 0.7 t = 0.3 t - 0.2$

Subtract $0.3 t$ from both sides

$0.4 {t}^{2} + 0.4 t = - 0.2$

Divide the equation by the leading coefficient of the ${t}^{2}$ term $0.4$.

${t}^{2} + t = - 0.5$

To complete the square, divide the coefficient of the $t$ term by $2$.

${t}^{2} + \textcolor{red}{1} t = - 0.5$

color(red)1 / 2 = color(blue)(0.5

Square $\textcolor{b l u e}{0.5}$ and add it to both sides of the equation.

${t}^{2} + t + {\left(\textcolor{b l u e}{0.5}\right)}^{2} = - 0.5 + {\left(\textcolor{b l u e}{0.5}\right)}^{2}$

Simplify

${t}^{2} + t + 0.25 = - 0.25$

Factor into a binomial squared. Note that the second term of the binomial is $\textcolor{b l u e}{0.25}$.

${\left(t + \textcolor{b l u e}{0.5}\right)}^{2} = - 0.25$

Square root both sides. Note that the negative sign in front of $- 0.25$ results in an $i$

$\sqrt{{\left(t + \textcolor{b l u e}{0.5}\right)}^{2}} = \sqrt{- 0.25}$

$t + 0.5 = \pm 0.5 i$

Subtract $0.5$ from both sides. Place it in front of the $\pm$.

$t = - 0.5 \pm 0.5 i$