# How do you solve 1/3t^2 + 3= 2t using the formula?

Jun 27, 2015

First subtract $2 t$ from both sides to get:

$\frac{1}{3} {t}^{2} - 2 t + 3 = 0$

$t = \frac{2 \pm \sqrt{{2}^{2} - \left(4 \times \frac{1}{3} \times 3\right)}}{2 \cdot \frac{1}{3}} = 3$

#### Explanation:

$\frac{1}{3} {t}^{2} - 2 t + 3$ is in the form $a {t}^{2} + b t + c$ with $a = \frac{1}{3}$, $b = - 2$ and $c = 3$

So the roots of $\frac{1}{3} {t}^{2} - 2 t + 3 = 0$ are given by the quadratic formula:

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{2 \pm \sqrt{{2}^{2} - \left(4 \times \frac{1}{3} \times 3\right)}}{2 \cdot \frac{1}{3}}$

$= \frac{2 \pm \sqrt{4 - 4}}{2 \cdot \frac{1}{3}}$

$= \frac{\cancel{2}}{\cancel{2} \cdot \frac{1}{3}}$

$= 3$

Alternatively, multiply the quadratic equation by $3$ to get:

${t}^{2} - 6 t + 9 = 0$

Notice that the coefficients (ignoring signs) are 1,6,9.

Does the pattern $1 , 6 , 9$ ring any bells?

Well $169 = {13}^{2}$ and non-coincidentally:

${t}^{2} - 6 t + 9 = {\left(t - 3\right)}^{2}$

${t}^{2} - 6 t + 9 = 0$ when $\left(t - 3\right) = 0$, that is when $t = 3$