Solve 4x^2-24=20x

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

Solve for x

$$x=-1$$

$$x=6$$

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.

$$a+b=-5 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

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

Calculate the sum for each pair.

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

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

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

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

Factor out x in x^{2}-6x.

$$x\left(x-6\right)+x-6$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+1\right)$$

To find equation solutions, solve x-6=0 and x+1=0.

$$x=6 x=-1$$

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}.

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

Square -20.

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

Multiply -4 times 4.

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

Multiply -16 times -24.

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

Add 400 to 384.

$$x=\frac{-\left(-20\right)±\sqrt{784}}{2\times 4}$$

Take the square root of 784.

$$x=\frac{-\left(-20\right)±28}{2\times 4}$$

The opposite of -20 is 20.

$$x=\frac{20±28}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{20±28}{8}$$

Now solve the equation x=\frac{20±28}{8} when ± is plus. Add 20 to 28.

$$x=\frac{48}{8}$$

Divide 48 by 8.

$$x=6$$

Now solve the equation x=\frac{20±28}{8} when ± is minus. Subtract 28 from 20.

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

Divide -8 by 8.

$$x=-1$$

The equation is now solved.

$$x=6 x=-1$$

Steps for Completing the Square

Subtract 20x from both sides.

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

Add 24 to both sides. Anything plus zero gives itself.

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

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}.

$$x^{2}-5x+\frac{25}{4}=\frac{49}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{7}{2} x-\frac{5}{2}=-\frac{7}{2}$$

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

$$x=6 x=-1$$