Solve 20x-10x^2

$$20 x - 10 x ^ { 2 }$$

Factor

$$10x\left(2-x\right)$$

Solution Steps

Factor out 10.

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

Consider 2x-x^{2}. Factor out x.

$$x\left(2-x\right)$$

Rewrite the complete factored expression.

$$10x\left(-x+2\right)$$

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.

$$-10x^{2}+20x=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}}}{2\left(-10\right)}$$

Take the square root of 20^{2}.

$$x=\frac{-20±20}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-20±20}{-20}$$

Now solve the equation x=\frac{-20±20}{-20} when ± is plus. Add -20 to 20.

$$x=\frac{0}{-20}$$

Divide 0 by -20.

$$x=0$$

Now solve the equation x=\frac{-20±20}{-20} when ± is minus. Subtract 20 from -20.

$$x=-\frac{40}{-20}$$

Divide -40 by -20.

$$x=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and 2 for x_{2}.

$$-10x^{2}+20x=-10x\left(x-2\right)$$

Evaluate

$$10x\left(2-x\right)$$