Solve (6-t)(16-2t)=16

$$( 6 - t ) ( 16 - 2 t ) = 16$$

Solve for t

$$t=4$$

$$t=10$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 6-t by 16-2t and combine like terms.

$$96-28t+2t^{2}=16$$

Subtract 16 from both sides.

$$96-28t+2t^{2}-16=0$$

Subtract 16 from 96 to get 80.

$$80-28t+2t^{2}=0$$

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.

$$2t^{2}-28t+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -28 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 2\times 80}}{2\times 2}$$

Square -28.

$$t=\frac{-\left(-28\right)±\sqrt{784-4\times 2\times 80}}{2\times 2}$$

Multiply -4 times 2.

$$t=\frac{-\left(-28\right)±\sqrt{784-8\times 80}}{2\times 2}$$

Multiply -8 times 80.

$$t=\frac{-\left(-28\right)±\sqrt{784-640}}{2\times 2}$$

Add 784 to -640.

$$t=\frac{-\left(-28\right)±\sqrt{144}}{2\times 2}$$

Take the square root of 144.

$$t=\frac{-\left(-28\right)±12}{2\times 2}$$

The opposite of -28 is 28.

$$t=\frac{28±12}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{28±12}{4}$$

Now solve the equation t=\frac{28±12}{4} when ± is plus. Add 28 to 12.

$$t=\frac{40}{4}$$

Divide 40 by 4.

$$t=10$$

Now solve the equation t=\frac{28±12}{4} when ± is minus. Subtract 12 from 28.

$$t=\frac{16}{4}$$

Divide 16 by 4.

$$t=4$$

The equation is now solved.

$$t=10 t=4$$

Steps for Completing the Square

Use the distributive property to multiply 6-t by 16-2t and combine like terms.

$$96-28t+2t^{2}=16$$

Subtract 96 from both sides.

$$-28t+2t^{2}=16-96$$

Subtract 96 from 16 to get -80.

$$-28t+2t^{2}=-80$$

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.

$$2t^{2}-28t=-80$$

Divide both sides by 2.

$$\frac{2t^{2}-28t}{2}=-\frac{80}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$t^{2}+\left(-\frac{28}{2}\right)t=-\frac{80}{2}$$

Divide -28 by 2.

$$t^{2}-14t=-\frac{80}{2}$$

Divide -80 by 2.

$$t^{2}-14t=-40$$

Divide -14, the coefficient of the x term, by 2 to get -7. Then add the square of -7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-14t+\left(-7\right)^{2}=-40+\left(-7\right)^{2}$$

Square -7.

$$t^{2}-14t+49=-40+49$$

Add -40 to 49.

$$t^{2}-14t+49=9$$

Factor t^{2}-14t+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-7\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-7\right)^{2}}=\sqrt{9}$$

Simplify.

$$t-7=3 t-7=-3$$

Add 7 to both sides of the equation.

$$t=10 t=4$$