Solve 2x^2-20x+50=0

$$2 { x }^{ 2 } -20x+50=0$$

Solve for x

$$x=5$$

Steps Using Factoring By Grouping

Divide both sides by 2.

$$x^{2}-10x+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 25=25$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.

$$-1,-25 -5,-5$$

Calculate the sum for each pair.

$$-1-25=-26 -5-5=-10$$

The solution is the pair that gives sum -10.

$$a=-5 b=-5$$

Rewrite x^{2}-10x+25 as \left(x^{2}-5x\right)+\left(-5x+25\right).

$$\left(x^{2}-5x\right)+\left(-5x+25\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-5\right)-5\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-5\right)$$

Rewrite as a binomial square.

$$\left(x-5\right)^{2}$$

To find equation solution, solve x-5=0.

$$x=5$$

Steps Using the Quadratic Formula

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.

$$2x^{2}-20x+50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -20 for b, and 50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 2\times 50}}{2\times 2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 2\times 50}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-20\right)±\sqrt{400-8\times 50}}{2\times 2}$$

Multiply -8 times 50.

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

Add 400 to -400.

$$x=\frac{-\left(-20\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$x=-\frac{-20}{2\times 2}$$

The opposite of -20 is 20.

$$x=\frac{20}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-20x+50=0$$

Subtract 50 from both sides of the equation.

$$2x^{2}-20x+50-50=-50$$

Subtracting 50 from itself leaves 0.

$$2x^{2}-20x=-50$$

Divide both sides by 2.

$$\frac{2x^{2}-20x}{2}=-\frac{50}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{20}{2}\right)x=-\frac{50}{2}$$

Divide -20 by 2.

$$x^{2}-10x=-\frac{50}{2}$$

Divide -50 by 2.

$$x^{2}-10x=-25$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-25+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-25+25$$

Add -25 to 25.

$$x^{2}-10x+25=0$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-5=0 x-5=0$$

Add 5 to both sides of the equation.

$$x=5 x=5$$

The equation is now solved. Solutions are the same.

$$x=5$$