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

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Answer 1 · 2017-11-27T15:17:11

$t = -0.5 +- 0.5i$

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

Subtract $0.3t$ from both sides

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

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 $2$ .

$t^2 + color(red)1t = -0.5$

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

Square $color(blue)(0.5)$ and add it to both sides of the equation.

$t^2 +t +(color(blue)(0.5))^2=-0.5 +(color(blue)(0.5))^2$

Simplify

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

Factor into a binomial squared. Note that the second term of the binomial is $color (blue)(0.25)$ .

$(t+color(blue)(0.5))^2 = -0.25$

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

$sqrt((t+color(blue)(0.5))^2) = sqrt(-0.25)$

$t + 0.5 = +-0.5i$

Subtract $0.5$ from both sides. Place it in front of the $+-$ .

$t=-0.5+-0.5i$

How do you solve 0.4t^2+0.7t=0.3t-0.20.4t^2+0.7t=0.3t-0.2 by completing the square?