# How do you solve  p^2+4p=-1?

Apr 12, 2016

$x = - 2 \pm \sqrt{3}$

#### Explanation:

color(blue)(p^2+4p=-1

Add $1$ both sides

$\rightarrow {p}^{2} + 4 p + 1 = - 1 + 1$

color(purple)(rarrp^2+4p+1=0

This is a Quadratic equation (in form $a {x}^{2} + b x + c = 0$)

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

Remember that $a \mathmr{and} b$ are the $\text{coefficients}$ of ${p}^{2} \mathmr{and} p$ and $c$ is the $\text{constant number}$

So,

$\textcolor{v i o \le t}{\left(a , b , c\right) = \left(1 , 4 , 1\right)}$

$\rightarrow x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(1\right) \left(1\right)}}{2 \left(1\right)}$

$\rightarrow x = \frac{- 4 \pm \sqrt{16 - 4}}{2}$

$\rightarrow x = \frac{- 4 \pm \sqrt{12}}{2}$

$\rightarrow x = \frac{- 4 \pm \sqrt{4 \cdot 3}}{2}$

$\rightarrow x = \frac{- 4 \pm 2 \sqrt{3}}{2}$

Cancel

$\rightarrow x = \frac{- {\cancel{4}}^{2} \pm {\cancel{2}}^{1} \sqrt{3}}{{\cancel{2}}^{1}}$

color(green)(x=-2+-sqrt3

Remember that the symbol $\pm$ means $\text{plus or minus}$

So,

It implies that,

color(indigo)(x=-2+sqrt3,-2-sqrt3

If you need more guidance for the Quadratic formula

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