# How do you solve (2x+3)(x+4)=1  for x using the square root property?

$\left(2 x + 3\right) \left(x + 4\right) = 1$
$2 {x}^{2} + 7 x + 12 = 1$
$2 \left({x}^{2} + \frac{7}{2} x + \frac{49}{16}\right) + 12 - \frac{49}{8} = 1$
$2 {\left(x + \frac{7}{4}\right)}^{2} + \frac{47}{8} = 1$
$2 {\left(x + \frac{7}{4}\right)}^{2} = - \frac{39}{8}$
$\left(x + \frac{7}{4}\right) = \setminus \pm \sqrt{\frac{39}{16}} i$
$x = \frac{- 7 \pm \sqrt{39} i}{4}$