Question

Solve$z^2+8z=-19$.

Short answer

The equation has no real solutions.

Step-by-step solution

Given Information  

The Given equation is$z^2+8z=-19$.

Isolate the variable terms on one side and the constant terms on the other. 

The variable terms are on the left side.

$z^2+8z=-19$

Here,$b=8$.

Find 12b2 and add to the both sides of equation.  

$$\begin{gathered} ={\left(\frac{1}{2}\left(8\right)\right)}^2 \\ ={\left(4\right)}^2 \\ =16 \end{gathered}$$

Add $16$both sides.

$$\begin{gathered} z^2+8z+16=-19+16 \\ {z}^2+8z+16=-3 \end{gathered}$$

Factor the perfect square trinomial as a binomial square.  

$(z+4{)}^2=-3$

Use the square root property.

$z+4=\pm \sqrt{-3}$

Since $\sqrt{-3}$is not a real number.

Thus, the equation has no real solutions.