Solve 2t=t+0.25t^2

$$2 t = t + 0.25 t ^ { 2 }$$

Solve for t

$$t=4$$

$$t=0$$

Steps Using Factoring

Subtract t from both sides.

$$2t-t=0.25t^{2}$$

Combine 2t and -t to get t.

$$t=0.25t^{2}$$

Subtract 0.25t^{2} from both sides.

$$t-0.25t^{2}=0$$

Factor out t.

$$t\left(1-0.25t\right)=0$$

To find equation solutions, solve t=0 and 1-\frac{t}{4}=0.

$$t=0 t=4$$

Steps Using the Quadratic Formula

Subtract t from both sides.

$$2t-t=0.25t^{2}$$

Combine 2t and -t to get t.

$$t=0.25t^{2}$$

Subtract 0.25t^{2} from both sides.

$$t-0.25t^{2}=0$$

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.

$$-0.25t^{2}+t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.25 for a, 1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-1±\sqrt{1^{2}}}{2\left(-0.25\right)}$$

Take the square root of 1^{2}.

$$t=\frac{-1±1}{2\left(-0.25\right)}$$

Multiply 2 times -0.25.

$$t=\frac{-1±1}{-0.5}$$

Now solve the equation t=\frac{-1±1}{-0.5} when ± is plus. Add -1 to 1.

$$t=\frac{0}{-0.5}$$

Divide 0 by -0.5 by multiplying 0 by the reciprocal of -0.5.

$$t=0$$

Now solve the equation t=\frac{-1±1}{-0.5} when ± is minus. Subtract 1 from -1.

$$t=-\frac{2}{-0.5}$$

Divide -2 by -0.5 by multiplying -2 by the reciprocal of -0.5.

$$t=4$$

The equation is now solved.

$$t=0 t=4$$

Steps for Completing the Square

Subtract t from both sides.

$$2t-t=0.25t^{2}$$

Combine 2t and -t to get t.

$$t=0.25t^{2}$$

Subtract 0.25t^{2} from both sides.

$$t-0.25t^{2}=0$$

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.

$$-0.25t^{2}+t=0$$

Multiply both sides by -4.

$$\frac{-0.25t^{2}+t}{-0.25}=\frac{0}{-0.25}$$

Dividing by -0.25 undoes the multiplication by -0.25.

$$t^{2}+\frac{1}{-0.25}t=\frac{0}{-0.25}$$

Divide 1 by -0.25 by multiplying 1 by the reciprocal of -0.25.

$$t^{2}-4t=\frac{0}{-0.25}$$

Divide 0 by -0.25 by multiplying 0 by the reciprocal of -0.25.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Factor t^{2}-4t+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$t-2=2 t-2=-2$$

Add 2 to both sides of the equation.

$$t=4 t=0$$