Solve (2x^2-4x+1)-3x

$$(2 { x }^{ 2 } -4x+1)-3x$$

Factor

$$2\left(x-\frac{7-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+7}{4}\right)$$

Steps Using the Quadratic Formula

Combine -4x and -3x to get -7x.

$$factor(2x^{2}-7x+1)$$

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.

$$2x^{2}-7x+1=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8}}{2\times 2}$$

Add 49 to -8.

$$x=\frac{-\left(-7\right)±\sqrt{41}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{41}}{4}$$

Now solve the equation x=\frac{7±\sqrt{41}}{4} when ± is plus. Add 7 to \sqrt{41}.

$$x=\frac{\sqrt{41}+7}{4}$$

Now solve the equation x=\frac{7±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from 7.

$$x=\frac{7-\sqrt{41}}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7+\sqrt{41}}{4} for x_{1} and \frac{7-\sqrt{41}}{4} for x_{2}.

$$2x^{2}-7x+1=2\left(x-\frac{\sqrt{41}+7}{4}\right)\left(x-\frac{7-\sqrt{41}}{4}\right)$$

Evaluate

$$2x^{2}-7x+1$$

Solution Steps

Combine -4x and -3x to get -7x.

$$2x^{2}-7x+1$$