Solve x^24-4(7x-12)=0

$${ x }^{ 2 } 4-4(7x-12)=0$$

Solve for x

$$x=3$$

$$x=4$$

Steps Using Factoring By Grouping

Use the distributive property to multiply -4 by 7x-12.

$$x^{2}\times 4-28x+48=0$$

Divide both sides by 4.

$$x^{2}-7x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 12=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite x^{2}-7x+12 as \left(x^{2}-4x\right)+\left(-3x+12\right).

$$\left(x^{2}-4x\right)+\left(-3x+12\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-4\right)-3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-3\right)$$

To find equation solutions, solve x-4=0 and x-3=0.

$$x=4 x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -4 by 7x-12.

$$x^{2}\times 4-28x+48=0$$

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.

$$4x^{2}-28x+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -28 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 4\times 48}}{2\times 4}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 4\times 48}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-28\right)±\sqrt{784-16\times 48}}{2\times 4}$$

Multiply -16 times 48.

$$x=\frac{-\left(-28\right)±\sqrt{784-768}}{2\times 4}$$

Add 784 to -768.

$$x=\frac{-\left(-28\right)±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

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

The opposite of -28 is 28.

$$x=\frac{28±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{28±4}{8}$$

Now solve the equation x=\frac{28±4}{8} when ± is plus. Add 28 to 4.

$$x=\frac{32}{8}$$

Divide 32 by 8.

$$x=4$$

Now solve the equation x=\frac{28±4}{8} when ± is minus. Subtract 4 from 28.

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

Divide 24 by 8.

$$x=3$$

The equation is now solved.

$$x=4 x=3$$

Steps for Completing the Square

Use the distributive property to multiply -4 by 7x-12.

$$x^{2}\times 4-28x+48=0$$

Subtract 48 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}\times 4-28x=-48$$

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.

$$4x^{2}-28x=-48$$

Divide both sides by 4.

$$\frac{4x^{2}-28x}{4}=-\frac{48}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{28}{4}\right)x=-\frac{48}{4}$$

Divide -28 by 4.

$$x^{2}-7x=-\frac{48}{4}$$

Divide -48 by 4.

$$x^{2}-7x=-12$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{1}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=4 x=3$$