Solve 4x^2+24=20x

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

Solve for x

$$x=2$$

$$x=3$$

Steps Using Factoring By Grouping

Subtract 20x from both sides.

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

Divide both sides by 4.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

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

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite x^{2}-5x+6 as \left(x^{2}-3x\right)+\left(-2x+6\right).

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

Factor out x in the first and -2 in the second group.

$$x\left(x-3\right)-2\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x-2\right)$$

To find equation solutions, solve x-3=0 and x-2=0.

$$x=3 x=2$$

Steps Using the Quadratic Formula

Subtract 20x from both sides.

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

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.

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

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

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

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 24}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 24}}{2\times 4}$$

Multiply -16 times 24.

$$x=\frac{-\left(-20\right)±\sqrt{400-384}}{2\times 4}$$

Add 400 to -384.

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

Take the square root of 16.

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

The opposite of -20 is 20.

$$x=\frac{20±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{20±4}{8}$$

Now solve the equation x=\frac{20±4}{8} when ± is plus. Add 20 to 4.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{20±4}{8} when ± is minus. Subtract 4 from 20.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

The equation is now solved.

$$x=3 x=2$$

Steps for Completing the Square

Subtract 20x from both sides.

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

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -20 by 4.

$$x^{2}-5x=-\frac{24}{4}$$

Divide -24 by 4.

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

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=3 x=2$$