# How do you solve  (x+1)^2-3=13?

Sep 6, 2016

$x = 3 \mathmr{and} x = - 5$

#### Explanation:

Although this is a quadratic equation, we do not have to use the normal method of making it equal to 0. This is a special case - there is no x-term.

Move all the constants to the right hand side and then find the square root of each side.

${\left(x + 1\right)}^{2} = 16$

$x + 1 = \pm \sqrt{16} = \pm 4$

$\mathmr{if} x + 1 = + 4 \text{ } \rightarrow x = 3$

$\mathmr{if} x + 1 = - 4 \text{ } \rightarrow x = - 5$