Question

which shows the correct substitution of the values a, b, and c from the equation 1 = -2x + 3x² + 1 into the quadratic formula?
quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
\\(\bigcirc\\) $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(0)}}{2(3)}$
\\(\bigcirc\\) $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(2)}}{2(3)}$
\\(\bigcirc\\) $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(1)}}{2(3)}$
\\(\bigcirc\\) $x = \frac{-3 \pm \sqrt{3^2 - 4(-2)(0)}}{2(-2)}$

Explanation:

Step1: Rewrite the equation in standard form

First, we need to rewrite the equation \(1 = -2x + 3x^2+1\) in the standard quadratic form \(ax^{2}+bx + c = 0\).
Subtract 1 from both sides of the equation:
\(1-1=-2x + 3x^{2}+1 - 1\)
Simplify to get:
\(0=3x^{2}-2x+0\)
So, the standard form is \(3x^{2}-2x + 0=0\), where \(a = 3\), \(b=- 2\) and \(c = 0\).

Step2: Substitute into the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)
Substitute \(a = 3\), \(b=-2\) and \(c = 0\) into the formula:
\(x=\frac{-(-2)\pm\sqrt{(-2)^{2}-4(3)(0)}}{2(3)}\)

Answer:

\(x=\frac{-(-2)\pm\sqrt{(-2)^{2}-4(3)(0)}}{2(3)}\) (the first option)