What are the solutions of [tex]$x^2 + 6x - 6 = 10$[/tex]?

A. [tex]x = -11[/tex] or [tex]x = 1[/tex]
B. [tex]x = -11[/tex] or [tex]x = -1[/tex]
C. [tex]x = -8[/tex] or [tex]x = -2[/tex]
D. [tex]x = -8[/tex] or [tex]x = 2[/tex]



Answer :

To find the solutions of the equation [tex]\( x^2 + 6x - 6 = 10 \)[/tex], we begin by rewriting it in standard form.

First, we subtract 10 from both sides:

[tex]\[ x^2 + 6x - 6 - 10 = 0 \][/tex]

This simplifies to:

[tex]\[ x^2 + 6x - 16 = 0 \][/tex]

Now we have a quadratic equation in the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -16 \)[/tex].

To solve this quadratic equation, we can use the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:

[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-16)}}{2(1)} \][/tex]

Simplify inside the square root:

[tex]\[ x = \frac{-6 \pm \sqrt{36 + 64}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{100}}{2} \][/tex]

Taking the square root of 100:

[tex]\[ x = \frac{-6 \pm 10}{2} \][/tex]

This gives us two possible solutions:

[tex]\[ x = \frac{-6 + 10}{2} = \frac{4}{2} = 2 \][/tex]

and

[tex]\[ x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8 \][/tex]

Therefore, the solutions to the equation [tex]\( x^2 + 6x - 6 = 10 \)[/tex] are:

[tex]\[ x = -8 \text{ or } x = 2 \][/tex]

So the correct choice is:

[tex]\[ x = -8 \text{ or } x = 2 \][/tex]