# How do you solve 4p^2=-7p-3 using the quadratic formula?

To solve, you must first move all the terms to one side, so: $4 {p}^{2} + 7 p + 3 = 0$. Then, the quadratic formula is $x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$. Plug in 4 for a, 7 for b and 3 for c. You get $x = \setminus \frac{- 7 \setminus \pm \setminus \sqrt{{7}^{2} - 4 \cdot 4 \cdot 3}}{2 \cdot 4} = \setminus \frac{- 7 \setminus \pm 1}{8}$ so, $x = \frac{- 7 + 1}{8} = - \frac{6}{8} = - \frac{3}{4}$ and $x = \frac{- 7 - 1}{8} = \frac{- 8}{8} = - 1$. x=-1 and -3/4