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

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Answer 1 · 2017-11-27T15:17:11

$t = - 0.5 \pm 0.5 i$ $t = -0.5 +- 0.5i$

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

Subtract $0.3 t$ $0.3t$ from both sides

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

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

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

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

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

$\frac{1}{2} = 0.5$ $color(red)1 / 2 = color(blue)(0.5$

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

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

Simplify

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

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

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

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

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

$t + 0.5 = \pm 0.5 i$ $t + 0.5 = +-0.5i$

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

$t = - 0.5 \pm 0.5 i$ $t=-0.5+-0.5i$

How do you solve 0.4t^2+0.7t=0.3t-0.20.4t2+0.7t=0.3t−0.20.4t^2+0.7t=0.3t-0.2 by completing the square?