Question

Use the quadratic formula to solve the equation. $$2 x^{2}-6 x+4=0$$

Short answer

The solutions to the equation are \(x = 1\) and \(x = 2\).

Step-by-step solution

Identify the Coefficients

In the quadratic equation \(2 x^{2}-6 x+4=0\), \(a\), \(b\), and \(c\) represent the coefficients as follows: \(a = 2\), \(b = -6\), and \(c = 4\).

Substitute into the Quadratic Formula

Now that we know the values of \(a\), \(b\), and \(c\), substitute these into the quadratic formula \(-\frac{b \pm \sqrt{b^{2}-4 a c}}{2 a}\). When you do that, you get \(-\frac{-6 \pm \sqrt{(-6)^{2}-4 \cdot 2 \cdot 4}}{2 \cdot 2}\).

Simplify Under The Square Root

Start by simplifying the expression under the square root sign. \((-6)^{2}-4 \cdot 2 \cdot 4 = 36 - 32 = 4.\) The expression now becomes \(-\frac{-6 \pm \sqrt{4}}{2 \cdot 2}\).

Simplify Further

The square root of 4 is 2, so the equation simplifies to \(-\frac{-6 \pm 2}{4}\). By applying the plus-minus, we get two possible solutions: \(-\frac{-6 + 2}{4}\) and \(-\frac{-6 - 2}{4}\). This further simplifies to \(-\frac{-4}{4}\) and \(-\frac{-8}{4}\).

Final Answer

Finally, simplifying \(-\frac{-4}{4}\) yields 1 and simplifying \(-\frac{-8}{4}\) yields 2. So, the solutions to the equation \(2x^2 - 6x + 4 = 0\) are \(x = 1\) and \(x = 2\).