Solve 2x^2-20x+200

$$2 { x }^{ 2 } -20x+200$$

Factor

$$2\left(x^{2}-10x+100\right)$$

Solution Steps

Factor out 2. Polynomial x^{2}-10x+100 is not factored since it does not have any rational roots.

$$2\left(x^{2}-10x+100\right)$$

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.

$$2x^{2}-20x+200=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 2\times 200}}{2\times 2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 2\times 200}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-20\right)±\sqrt{400-8\times 200}}{2\times 2}$$

Multiply -8 times 200.

$$x=\frac{-\left(-20\right)±\sqrt{400-1600}}{2\times 2}$$

Add 400 to -1600.

$$x=\frac{-\left(-20\right)±\sqrt{-1200}}{2\times 2}$$

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

$$2x^{2}-20x+200$$

Evaluate

$$2\left(x^{2}-10x+100\right)$$