# 3x2 -6x - 4 =0 how to complete the square?

Feb 16, 2018

See below...

#### Explanation:

We have the quadratic $3 {x}^{2} - 6 x - 4 = 0$

First of all, we take out a factor of 3. Don't take it out from the constant however as it can lead to some unnecessary fraction work.

$3 {x}^{2} - 6 x - 4 \implies 3 \left[{x}^{2} - 2 x\right] - 4$

Now we write out our initial bracket. To do this we have

${\left(x + \frac{b}{2}\right)}^{2}$ $\implies$ in this case $b$ is $- 2$. Note that we don't include an $x$ after the $b$...
Once we have our initial bracket, we subtract the square of $\frac{b}{2}$

$\therefore 3 \left[{x}^{2} - 2 x\right] - 4 \implies 3 \left[{\left(x - 1\right)}^{2} - 1\right] - 4$

Now we must remove the square brackets by multiplying whats in it by the factor on the outside, in this case $3$.

$\therefore$ we get $3 {\left(x - 1\right)}^{2} - 3 - 4 = 3 {\left(x - 1\right)}^{2} - 7$

Final answer $3 {\left(x - 1\right)}^{2} - 7 = 0$