Answer:
x1 = -1.59
x2 = -4.41
Step-by-step explanation:
The quadratic equation is of the form
\(ax^{2}+bx+c=0\)
In the given case, b = 6
We apply the formula
\((\frac{b}{2})^{2}=(\frac{6}{2})^{2}=3^{2}=9\)
\(x^{2}+6x+7+9-9=0\)
\(x^{2}+6x+9+7-9=0\)
\((x^{2}+6x+9)-2=0\)
\((x+3)(x-3)=2\)
\((x+3)^{2}=2\)
Applying square root to both parts of the equation, we get,
\(\sqrt{(x+3)^{2}} =\sqrt{2}\)
\(x+3=\sqrt{2}\)
\(x=\sqrt{2}-3\)
x = ± 1.41 - 3
"x" then has two values:
x1 = 1.41 - 3 = -1.59
x2 = -1.41 - 3 = -4.41
Hope this helps!
