Solve 4x^2-24x+20=0

$$4 { x }^{ 2 } -24x+20=0$$

Solve for x

$$x=1$$

$$x=5$$

Steps Using Factoring By Grouping

Divide both sides by 4.

$$x^{2}-6x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

Rewrite x^{2}-6x+5 as \left(x^{2}-5x\right)+\left(-x+5\right).

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

Factor out x in the first and -1 in the second group.

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

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

$$\left(x-5\right)\left(x-1\right)$$

To find equation solutions, solve x-5=0 and x-1=0.

$$x=5 x=1$$

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.

$$4x^{2}-24x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -24 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 4\times 20}}{2\times 4}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 4\times 20}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-24\right)±\sqrt{576-16\times 20}}{2\times 4}$$

Multiply -16 times 20.

$$x=\frac{-\left(-24\right)±\sqrt{576-320}}{2\times 4}$$

Add 576 to -320.

$$x=\frac{-\left(-24\right)±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

$$x=\frac{-\left(-24\right)±16}{2\times 4}$$

The opposite of -24 is 24.

$$x=\frac{24±16}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{24±16}{8}$$

Now solve the equation x=\frac{24±16}{8} when ± is plus. Add 24 to 16.

$$x=\frac{40}{8}$$

Divide 40 by 8.

$$x=5$$

Now solve the equation x=\frac{24±16}{8} when ± is minus. Subtract 16 from 24.

$$x=\frac{8}{8}$$

Divide 8 by 8.

$$x=1$$

The equation is now solved.

$$x=5 x=1$$

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.

$$4x^{2}-24x+20=0$$

Subtract 20 from both sides of the equation.

$$4x^{2}-24x+20-20=-20$$

Subtracting 20 from itself leaves 0.

$$4x^{2}-24x=-20$$

Divide both sides by 4.

$$\frac{4x^{2}-24x}{4}=-\frac{20}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{24}{4}\right)x=-\frac{20}{4}$$

Divide -24 by 4.

$$x^{2}-6x=-\frac{20}{4}$$

Divide -20 by 4.

$$x^{2}-6x=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-5+9$$

Add -5 to 9.

$$x^{2}-6x+9=4$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$