Certainly! Let's solve the given quadratic equation step by step.
The given equation is:
[tex]\[ 6x^2 + 6 = 12x + 18 \][/tex]
First, we need to rewrite the equation in standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex].
Step 1: Move all terms to one side of the equation to set it equal to zero:
[tex]\[ 6x^2 + 6 - 12x - 18 = 0 \][/tex]
Step 2: Simplify the equation by combining like terms:
[tex]\[ 6x^2 - 12x + 6 - 18 = 0 \][/tex]
[tex]\[ 6x^2 - 12x - 12 = 0 \][/tex]
Step 3: Divide the entire equation by 6 to simplify it:
[tex]\[ x^2 - 2x - 2 = 0 \][/tex]
Now, we have a quadratic equation in the standard form:
[tex]\[ x^2 - 2x - 2 = 0 \][/tex]
Step 4: To find the solutions, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -2 \)[/tex].
Step 5: Calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-2)^2 - 4(1)(-2) \][/tex]
[tex]\[ \Delta = 4 + 8 \][/tex]
[tex]\[ \Delta = 12 \][/tex]
Step 6: Substitute the discriminant back into the quadratic formula to find the two solutions:
[tex]\[ x = \frac{-(-2) \pm \sqrt{12}}{2(1)} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{12}}{2} \][/tex]
[tex]\[ \sqrt{12} \][/tex] can be simplified to [tex]\( 2\sqrt{3} \)[/tex], so:
[tex]\[ x = \frac{2 \pm 2\sqrt{3}}{2} \][/tex]
[tex]\[ x = 1 \pm \sqrt{3} \][/tex]
Therefore, the solutions to the quadratic equation are:
[tex]\[ x = 1 + \sqrt{3} \][/tex]
[tex]\[ x = 1 - \sqrt{3} \][/tex]
The correct answer is:
A. [tex]\( x = 1 \pm \sqrt{3} \)[/tex]
