Solve 4t^2-10t=0

$$4 t ^ { 2 } - 10 t = 0$$

Solve for t

$$t = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$t=0$$

Steps Using Factoring

Factor out t.

$$t\left(4t-10\right)=0$$

To find equation solutions, solve t=0 and 4t-10=0.

$$t=0 t=\frac{5}{2}$$

Steps Using the Quadratic Formula

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.

$$4t^{2}-10t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2\times 4}$$

Take the square root of \left(-10\right)^{2}.

$$t=\frac{-\left(-10\right)±10}{2\times 4}$$

The opposite of -10 is 10.

$$t=\frac{10±10}{2\times 4}$$

Multiply 2 times 4.

$$t=\frac{10±10}{8}$$

Now solve the equation t=\frac{10±10}{8} when ± is plus. Add 10 to 10.

$$t=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$t=\frac{5}{2}$$

Now solve the equation t=\frac{10±10}{8} when ± is minus. Subtract 10 from 10.

$$t=\frac{0}{8}$$

Divide 0 by 8.

$$t=0$$

The equation is now solved.

$$t=\frac{5}{2} t=0$$

Steps for Completing the Square

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.

$$4t^{2}-10t=0$$

Divide both sides by 4.

$$\frac{4t^{2}-10t}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$t^{2}+\left(-\frac{10}{4}\right)t=\frac{0}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

$$t^{2}-\frac{5}{2}t=\frac{0}{4}$$

Divide 0 by 4.

$$t^{2}-\frac{5}{2}t=0$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{5}{2}t+\left(-\frac{5}{4}\right)^{2}=\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{5}{2}t+\frac{25}{16}=\frac{25}{16}$$

Factor t^{2}-\frac{5}{2}t+\frac{25}{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}.

$$\left(t-\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

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

Add \frac{5}{4} to both sides of the equation.

$$t=\frac{5}{2} t=0$$