Solve 4x^2-32x+60=12

$$4 { x }^{ 2 } -32x+60=12$$

Solve for x

$$x=2$$

$$x=6$$

Steps Using Factoring By Grouping

Subtract 12 from both sides.

$$4x^{2}-32x+60-12=0$$

Subtract 12 from 60 to get 48.

$$4x^{2}-32x+48=0$$

Divide both sides by 4.

$$x^{2}-8x+12=0$$

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.

$$a+b=-8 ab=1\times 12=12$$

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 -8.

$$a=-6 b=-2$$

Rewrite x^{2}-8x+12 as \left(x^{2}-6x\right)+\left(-2x+12\right).

$$\left(x^{2}-6x\right)+\left(-2x+12\right)$$

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

$$x\left(x-6\right)-2\left(x-6\right)$$

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

$$\left(x-6\right)\left(x-2\right)$$

To find equation solutions, solve x-6=0 and x-2=0.

$$x=6 x=2$$

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.

$$4x^{2}-32x+60=12$$

Subtract 12 from both sides of the equation.

$$4x^{2}-32x+60-12=12-12$$

Subtracting 12 from itself leaves 0.

$$4x^{2}-32x+60-12=0$$

Subtract 12 from 60.

$$4x^{2}-32x+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -32 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 4\times 48}}{2\times 4}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 4\times 48}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-32\right)±\sqrt{1024-16\times 48}}{2\times 4}$$

Multiply -16 times 48.

$$x=\frac{-\left(-32\right)±\sqrt{1024-768}}{2\times 4}$$

Add 1024 to -768.

$$x=\frac{-\left(-32\right)±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

$$x=\frac{-\left(-32\right)±16}{2\times 4}$$

The opposite of -32 is 32.

$$x=\frac{32±16}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{32±16}{8}$$

Now solve the equation x=\frac{32±16}{8} when ± is plus. Add 32 to 16.

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

Divide 48 by 8.

$$x=6$$

Now solve the equation x=\frac{32±16}{8} when ± is minus. Subtract 16 from 32.

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

Divide 16 by 8.

$$x=2$$

The equation is now solved.

$$x=6 x=2$$

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.

$$4x^{2}-32x+60=12$$

Subtract 60 from both sides of the equation.

$$4x^{2}-32x+60-60=12-60$$

Subtracting 60 from itself leaves 0.

$$4x^{2}-32x=12-60$$

Subtract 60 from 12.

$$4x^{2}-32x=-48$$

Divide both sides by 4.

$$\frac{4x^{2}-32x}{4}=-\frac{48}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{32}{4}\right)x=-\frac{48}{4}$$

Divide -32 by 4.

$$x^{2}-8x=-\frac{48}{4}$$

Divide -48 by 4.

$$x^{2}-8x=-12$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-12+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-12+16$$

Add -12 to 16.

$$x^{2}-8x+16=4$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=2 x-4=-2$$

Add 4 to both sides of the equation.

$$x=6 x=2$$