Solve 10-t=t/8t

$$10 - t = \frac { t } { 8 } t$$

Solve for t

$$t=4\sqrt{6}-4\approx 5.797958971$$

$$t=-4\sqrt{6}-4\approx -13.797958971$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 8.

$$80-8t=tt$$

Multiply t and t to get t^{2}.

$$80-8t=t^{2}$$

Subtract t^{2} from both sides.

$$80-8t-t^{2}=0$$

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.

$$-t^{2}-8t+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -8 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-1\right)\times 80}}{2\left(-1\right)}$$

Square -8.

$$t=\frac{-\left(-8\right)±\sqrt{64-4\left(-1\right)\times 80}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-\left(-8\right)±\sqrt{64+4\times 80}}{2\left(-1\right)}$$

Multiply 4 times 80.

$$t=\frac{-\left(-8\right)±\sqrt{64+320}}{2\left(-1\right)}$$

Add 64 to 320.

$$t=\frac{-\left(-8\right)±\sqrt{384}}{2\left(-1\right)}$$

Take the square root of 384.

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

The opposite of -8 is 8.

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

Multiply 2 times -1.

$$t=\frac{8±8\sqrt{6}}{-2}$$

Now solve the equation t=\frac{8±8\sqrt{6}}{-2} when ± is plus. Add 8 to 8\sqrt{6}.

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

Divide 8+8\sqrt{6} by -2.

$$t=-4\sqrt{6}-4$$

Now solve the equation t=\frac{8±8\sqrt{6}}{-2} when ± is minus. Subtract 8\sqrt{6} from 8.

$$t=\frac{8-8\sqrt{6}}{-2}$$

Divide 8-8\sqrt{6} by -2.

$$t=4\sqrt{6}-4$$

The equation is now solved.

$$t=-4\sqrt{6}-4 t=4\sqrt{6}-4$$

Steps for Completing the Square

Multiply both sides of the equation by 8.

$$80-8t=tt$$

Multiply t and t to get t^{2}.

$$80-8t=t^{2}$$

Subtract t^{2} from both sides.

$$80-8t-t^{2}=0$$

Subtract 80 from both sides. Anything subtracted from zero gives its negation.

$$-8t-t^{2}=-80$$

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.

$$-t^{2}-8t=-80$$

Divide both sides by -1.

$$\frac{-t^{2}-8t}{-1}=-\frac{80}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\left(-\frac{8}{-1}\right)t=-\frac{80}{-1}$$

Divide -8 by -1.

$$t^{2}+8t=-\frac{80}{-1}$$

Divide -80 by -1.

$$t^{2}+8t=80$$

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.

$$t^{2}+8t+4^{2}=80+4^{2}$$

Square 4.

$$t^{2}+8t+16=80+16$$

Add 80 to 16.

$$t^{2}+8t+16=96$$

Factor t^{2}+8t+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}.

$$\left(t+4\right)^{2}=96$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+4\right)^{2}}=\sqrt{96}$$

Simplify.

$$t+4=4\sqrt{6} t+4=-4\sqrt{6}$$

Subtract 4 from both sides of the equation.

$$t=4\sqrt{6}-4 t=-4\sqrt{6}-4$$