Answer:
[tex] x = -2 [/tex] or [tex] x = -5 [/tex]
Step-by-step explanation:
You need to complete the square before you can take the square root of both sides.
[tex] x^2 + 7x + 10 = 0 [/tex]
Subtract 10 from both sides.
[tex] x^2 + 7x = -10 [/tex]
To complete the square, you need to add the square of half of the x-term coefficient to both sides.
The x-term coefficient is 7. Half of that is 7/2. Square it to get 49/4. Now we add 49/4 to both sides of the equation.
[tex] x^2 + 7x + \dfrac{49}{4} = -10 + \dfrac{49}{4} [/tex]
[tex] (x + \dfrac{7}{2})^2 = -\dfrac{40}{4} + \dfrac{49}{4} [/tex]
[tex] (x + \dfrac{7}{2})^2 = \dfrac{9}{4} [/tex]
Now we use the square root property, if
[tex] x^2 = k [/tex], then
[tex] x = \pm \sqrt{k} [/tex]
[tex] x + \dfrac{7}{2} = \pm \sqrt{\dfrac{9}{4}} [/tex]
[tex] x + \dfrac{7}{2} = \pm \dfrac{3}{2} [/tex]
[tex] x + \dfrac{7}{2} = \dfrac{3}{2} [/tex] or [tex] x + \dfrac{7}{2} = -\dfrac{3}{2} [/tex]
[tex] x = -\dfrac{4}{2} [/tex] or [tex] x = -\dfrac{10}{2} [/tex]
[tex] x = -2 [/tex] or [tex] x = -5 [/tex]
