# How do you solve x^2+3-4x=0?

Jul 28, 2016

$x = 3 \mathmr{and} x = 1$

#### Explanation:

The first approach with quadratic equations is to see if we can find factors. Use $a {x}^{2} + b x + c = 0$

${x}^{2} - 4 x + 3 = 0$

Find factors of 3 which add up to 4.
The signs are the same, both minus.

$\left(x - 3\right) \left(x - 1\right) = 0$

Either of the factors can be 0.so:
$x = 3 \mathmr{and} x = 1$

Jul 29, 2016

$y = {x}^{2} - 4 x + 3 = 0$
Since a + b + c = 0, use shortcut. One real root is 1 and the other is $\frac{c}{a} = 3$