# How do you solve using completing the square method -3x^2+30x-74?

##### 2 Answers
Aug 5, 2016

This cannot really be "solved" because it is not an equation and it is not "equal" to anything. It can only be written in a different form.

Aug 5, 2016

I got:

$y = - 3 {\left(x - 5\right)}^{2} + 1$

You have:

$y = - 3 {x}^{2} + 30 x - 74$

What you can start with is to make the ${x}^{2}$ term have a perfect-square coefficient ($1$ is fine). That makes it easier to come up with a factor.

Then, halve the $x$ term's coefficient and square it, adding it to both sides.

$\frac{y}{3} = - {x}^{2} + 10 x - \frac{74}{3}$ (divide by 3)

$- \frac{y}{3} = {x}^{2} - 10 x + \frac{74}{3}$ (switch signs)

Half of $- 10$ is $- 5$, squared is $+ 25$. To make common denominators happen, $25 = \frac{75}{3}$. Now add it to both sides.

$\implies - \frac{y}{3} + \frac{75}{3} = {x}^{2} - 10 x + \frac{74}{3} + \frac{75}{3}$

$\implies - \frac{y}{3} = {x}^{2} - 10 x + \frac{75}{3} - \frac{1}{3}$

$\implies - \frac{y}{3} = \textcolor{g r e e n}{{x}^{2} - 10 x + 25} - \frac{1}{3}$

$\implies - \frac{y}{3} = \textcolor{g r e e n}{{\left(x - 5\right)}^{2}} - \frac{1}{3}$

Now you can return it back to $y$ by reversing the two things we did in the first steps:

$\implies \frac{y}{3} = - {\left(x - 5\right)}^{2} + \frac{1}{3}$

$\implies \textcolor{b l u e}{y = - 3 {\left(x - 5\right)}^{2} + 1}$