Solve 4-2t/2t=2t/4

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

Solve for t

$$t=\sqrt{5}-1\approx 1.236067977$$

$$t=-\sqrt{5}-1\approx -3.236067977$$

Steps Using the Quadratic Formula

Variable t cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4t, the least common multiple of 2t,4.

$$2\left(4-2t\right)=t\times 2t$$

Use the distributive property to multiply 2 by 4-2t.

$$8-4t=t\times 2t$$

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

$$8-4t=t^{2}\times 2$$

Subtract t^{2}\times 2 from both sides.

$$8-4t-t^{2}\times 2=0$$

Multiply -1 and 2 to get -2.

$$8-4t-2t^{2}=0$$

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.

$$-2t^{2}-4t+8=0$$

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

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

Square -4.

$$t=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$t=\frac{-\left(-4\right)±\sqrt{16+8\times 8}}{2\left(-2\right)}$$

Multiply 8 times 8.

$$t=\frac{-\left(-4\right)±\sqrt{16+64}}{2\left(-2\right)}$$

Add 16 to 64.

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

Take the square root of 80.

$$t=\frac{-\left(-4\right)±4\sqrt{5}}{2\left(-2\right)}$$

The opposite of -4 is 4.

$$t=\frac{4±4\sqrt{5}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$t=\frac{4±4\sqrt{5}}{-4}$$

Now solve the equation t=\frac{4±4\sqrt{5}}{-4} when ± is plus. Add 4 to 4\sqrt{5}.

$$t=\frac{4\sqrt{5}+4}{-4}$$

Divide 4+4\sqrt{5} by -4.

$$t=-\left(\sqrt{5}+1\right)$$

Now solve the equation t=\frac{4±4\sqrt{5}}{-4} when ± is minus. Subtract 4\sqrt{5} from 4.

$$t=\frac{4-4\sqrt{5}}{-4}$$

Divide 4-4\sqrt{5} by -4.

$$t=\sqrt{5}-1$$

The equation is now solved.

$$t=-\left(\sqrt{5}+1\right) t=\sqrt{5}-1$$

Steps for Completing the Square

Variable t cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4t, the least common multiple of 2t,4.

$$2\left(4-2t\right)=t\times 2t$$

Use the distributive property to multiply 2 by 4-2t.

$$8-4t=t\times 2t$$

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

$$8-4t=t^{2}\times 2$$

Subtract t^{2}\times 2 from both sides.

$$8-4t-t^{2}\times 2=0$$

Multiply -1 and 2 to get -2.

$$8-4t-2t^{2}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$-4t-2t^{2}=-8$$

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.

$$-2t^{2}-4t=-8$$

Divide both sides by -2.

$$\frac{-2t^{2}-4t}{-2}=-\frac{8}{-2}$$

Dividing by -2 undoes the multiplication by -2.

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

Divide -4 by -2.

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

Divide -8 by -2.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+2t+1^{2}=4+1^{2}$$

Square 1.

$$t^{2}+2t+1=4+1$$

Add 4 to 1.

$$t^{2}+2t+1=5$$

Factor t^{2}+2t+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+1\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+1\right)^{2}}=\sqrt{5}$$

Simplify.

$$t+1=\sqrt{5} t+1=-\sqrt{5}$$

Subtract 1 from both sides of the equation.

$$t=\sqrt{5}-1 t=-\sqrt{5}-1$$