To solve the given quadratic equation using the quadratic formula, we first need to rewrite the equation in standard form: [tex]\( ax^2 + bx + c = 0 \)[/tex].
Given equation:
[tex]\[ x^2 + 1 = 2x - 3 \][/tex]
Subtract [tex]\( 2x \)[/tex] and add [tex]\( 3 \)[/tex] to both sides to rewrite it in standard form:
[tex]\[ x^2 - 2x + 1 + 3 = 0 \][/tex]
[tex]\[ x^2 - 2x + 4 = 0 \][/tex]
Now, the equation is in standard form with:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = -2 \][/tex]
[tex]\[ c = 4 \][/tex]
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Now we apply the values [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = 4 \)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(4)}}{2(1)} \][/tex]
Simplifying inside the square root:
[tex]\[ x = \frac{2 \pm \sqrt{4 - 16}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{-12}}{2} \][/tex]
Now let's examine the given answer choices:
A:
[tex]\[ \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(4)}}{2(1)} \][/tex]
[tex]\[ = \frac{2 \pm \sqrt{4 - 16}}{2} \][/tex]
[tex]\[ = \frac{2 \pm \sqrt{-12}}{2} \][/tex]
B:
[tex]\[ \frac{-(-2) \pm \sqrt{(-2)^2 - (1)(4)}}{2(2)} \][/tex]
[tex]\[ = \frac{2 \pm \sqrt{4 - 4}}{4} \][/tex]
[tex]\[ = \frac{2 \pm \sqrt{0}}{4} \][/tex]
[tex]\[ = \frac{2 \pm 0}{4} \][/tex]
[tex]\[ = \frac{2}{4} \][/tex]
[tex]\[ = \frac{1}{2} \][/tex]
C:
[tex]\[ \frac{-2 \pm \sqrt{(-2)^2 - 4(1)(4)}}{2(1)} \][/tex]
[tex]\[ = \frac{-2 \pm \sqrt{4 - 16}}{2} \][/tex]
[tex]\[ = \frac{-2 \pm \sqrt{-12}}{2} \][/tex]
D:
[tex]\[ \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-2)}}{2(1)} \][/tex]
[tex]\[ = \frac{-2 \pm \sqrt{4 + 8}}{2} \][/tex]
[tex]\[ = \frac{-2 \pm \sqrt{12}}{2} \][/tex]
According to the quadratic formula, the correct setup among the given choices is:
[tex]\[ C. \frac{-2 \pm \sqrt{(-2)^2 - 4(1)(4)}}{2(1)} \][/tex]
So, the correct answer is [tex]\( \boxed{C} \)[/tex].
