First we need to factor trinomial $ \color{blue}{ -x^2-9x-8 } $ and than we use factored form to solve an equation $ \color{blue}{ -x^2-9x-8 = 0} $.

Step 1: We can simplify equation by multiplying both sides by -1. After multiplying we have the following equation:

$$ \begin{aligned} -x^2-9x-8 &= 0 \,\,\, / \color{orangered}{\cdot \, -1 } \\[0.9 em ] x^2+9x+8 &=0 \end{aligned} $$

Step 1: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = 9 } ~ \text{ and } ~ \color{red}{ c = 8 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 9 } $ and multiply to $ \color{red}{ 8 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 8 }$.

PRODUCT = 8
1     8	-1     -8
2     4	-2     -4

Step 3: Find out which pair sums up to $\color{blue}{ b = 9 }$

PRODUCT = 8 and SUM = 9
1     8	-1     -8
2     4	-2     -4

Step 4: Put 1 and 8 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+9x+8 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+9x+8 & = (x + 1)(x + 8) \end{aligned} $$

Step 5: Set each factor to zero and solve equations.

$$ \begin{array}{ccc} \begin{aligned} x+1 &= 0 \\ x &= -1 \end{aligned} & ~ & \begin{aligned} x+8 &= 0 \\ x &= -8 \end{aligned} \end{array} $$

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