Solve 2x^2-6x+6

$$2 { x }^{ 2 } -6x+6$$

Factor

$$2\left(x^{2}-3x+3\right)$$

Solution Steps

Factor out 2. Polynomial x^{2}-3x+3 is not factored since it does not have any rational roots.

$$2\left(x^{2}-3x+3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$2x^{2}-6x+6=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-\left(-6\right)±\sqrt{36-48}}{2\times 2}$$

Add 36 to -48.

$$x=\frac{-\left(-6\right)±\sqrt{-12}}{2\times 2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$2x^{2}-6x+6$$

Evaluate

$$2\left(x^{2}-3x+3\right)$$