Solve t^2-12=4t

$$t ^ { 2 } - 12 = 4 t$$

Solve for t

$$t=-2$$

$$t=6$$

Steps Using Factoring

Subtract 4t from both sides.

$$t^{2}-12-4t=0$$

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

$$t^{2}-4t-12=0$$

To solve the equation, factor t^{2}-4t-12 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.

$$a+b=-4 ab=-12$$

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 -12.

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

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.

$$\left(t-6\right)\left(t+2\right)$$

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

$$t=6 t=-2$$

Steps Using Factoring By Grouping

Subtract 4t from both sides.

$$t^{2}-12-4t=0$$

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

$$t^{2}-4t-12=0$$

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

$$a+b=-4 ab=1\left(-12\right)=-12$$

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 -12.

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

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite t^{2}-4t-12 as \left(t^{2}-6t\right)+\left(2t-12\right).

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

Factor out t in the first and 2 in the second group.

$$t\left(t-6\right)+2\left(t-6\right)$$

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

$$\left(t-6\right)\left(t+2\right)$$

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

$$t=6 t=-2$$

Steps Using the Quadratic Formula

Subtract 4t from both sides.

$$t^{2}-12-4t=0$$

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.

$$t^{2}-4t-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-12\right)}}{2}$$

Square -4.

$$t=\frac{-\left(-4\right)±\sqrt{16-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$t=\frac{-\left(-4\right)±\sqrt{16+48}}{2}$$

Add 16 to 48.

$$t=\frac{-\left(-4\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$t=\frac{-\left(-4\right)±8}{2}$$

The opposite of -4 is 4.

$$t=\frac{4±8}{2}$$

Now solve the equation t=\frac{4±8}{2} when ± is plus. Add 4 to 8.

$$t=\frac{12}{2}$$

Divide 12 by 2.

$$t=6$$

Now solve the equation t=\frac{4±8}{2} when ± is minus. Subtract 8 from 4.

$$t=-\frac{4}{2}$$

Divide -4 by 2.

$$t=-2$$

The equation is now solved.

$$t=6 t=-2$$

Steps for Completing the Square

Subtract 4t from both sides.

$$t^{2}-12-4t=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$t^{2}-4t=12$$

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

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

Square -2.

$$t^{2}-4t+4=12+4$$

Add 12 to 4.

$$t^{2}-4t+4=16$$

Factor t^{2}-4t+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}.

$$\left(t-2\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-2\right)^{2}}=\sqrt{16}$$

Simplify.

$$t-2=4 t-2=-4$$

Add 2 to both sides of the equation.

$$t=6 t=-2$$