First we need to factor trinomial $ \color{blue}{ -x^2+8x-12 } $ and than we use factored form to solve an equation $ \color{blue}{ -x^2+8x-12 = 0} $.
Step 1: We can simplify equation by multiplying both sides by -1. After multiplying we have the following equation:
$$ \begin{aligned} -x^2+8x-12 &= 0 \,\,\, / \color{orangered}{\cdot \, -1 } \\[0.9 em ] x^2-8x+12 &=0 \end{aligned} $$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:
$$ \color{blue}{ b = -8 } ~ \text{ and } ~ \color{red}{ c = 12 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -8 } $ and multiply to $ \color{red}{ 12 } $.
Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 12 }$.
PRODUCT = 12 | |
1 12 | -1 -12 |
2 6 | -2 -6 |
3 4 | -3 -4 |
Step 3: Find out which pair sums up to $\color{blue}{ b = -8 }$
PRODUCT = 12 and SUM = -8 | |
1 12 | -1 -12 |
2 6 | -2 -6 |
3 4 | -3 -4 |
Step 4: Put -2 and -6 into placeholders to get factored form.
$$ \begin{aligned} x^{2}-8x+12 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-8x+12 & = (x -2)(x -6) \end{aligned} $$Step 5: Set each factor to zero and solve equations.
$$ \begin{array}{ccc} \begin{aligned} x-2 &= 0 \\ x &= 2 \end{aligned} & ~ & \begin{aligned} x-6 &= 0 \\ x &= 6 \end{aligned} \end{array} $$