Solve 8=6t-t^2

$$8 = 6 t - t ^ { 2 }$$

Solve for t

$$t=2$$

$$t=4$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$6t-t^{2}=8$$

Subtract 8 from both sides.

$$6t-t^{2}-8=0$$

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

$$-t^{2}+6t-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt-8. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-8\right)=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=4 b=2$$

Rewrite -t^{2}+6t-8 as \left(-t^{2}+4t\right)+\left(2t-8\right).

$$\left(-t^{2}+4t\right)+\left(2t-8\right)$$

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

$$-t\left(t-4\right)+2\left(t-4\right)$$

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

$$\left(t-4\right)\left(-t+2\right)$$

To find equation solutions, solve t-4=0 and -t+2=0.

$$t=4 t=2$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$6t-t^{2}=8$$

Subtract 8 from both sides.

$$6t-t^{2}-8=0$$

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.

$$-t^{2}+6t-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Square 6.

$$t=\frac{-6±\sqrt{36-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-6±\sqrt{36+4\left(-8\right)}}{2\left(-1\right)}$$

Multiply 4 times -8.

$$t=\frac{-6±\sqrt{36-32}}{2\left(-1\right)}$$

Add 36 to -32.

$$t=\frac{-6±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$t=\frac{-6±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-6±2}{-2}$$

Now solve the equation t=\frac{-6±2}{-2} when ± is plus. Add -6 to 2.

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

Divide -4 by -2.

$$t=2$$

Now solve the equation t=\frac{-6±2}{-2} when ± is minus. Subtract 2 from -6.

$$t=-\frac{8}{-2}$$

Divide -8 by -2.

$$t=4$$

The equation is now solved.

$$t=2 t=4$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$6t-t^{2}=8$$

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.

$$-t^{2}+6t=8$$

Divide both sides by -1.

$$\frac{-t^{2}+6t}{-1}=\frac{8}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\frac{6}{-1}t=\frac{8}{-1}$$

Divide 6 by -1.

$$t^{2}-6t=\frac{8}{-1}$$

Divide 8 by -1.

$$t^{2}-6t=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-6t+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$t^{2}-6t+9=-8+9$$

Add -8 to 9.

$$t^{2}-6t+9=1$$

Factor t^{2}-6t+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-3\right)^{2}}=\sqrt{1}$$

Simplify.

$$t-3=1 t-3=-1$$

Add 3 to both sides of the equation.

$$t=4 t=2$$