Solve 2tleft(6-tright)1/2=8

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

Solve for t

$$t=2$$

$$t=4$$

Steps Using the Quadratic Formula

Cancel out 2 and 2.

$$t\left(6-t\right)=8$$

Use the distributive property to multiply t by 6-t.

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

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

Cancel out 2 and 2.

$$t\left(6-t\right)=8$$

Use the distributive property to multiply t by 6-t.

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

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.

$$t=4 t=2$$