Solve (5-t)t=4

$$( 5 - t ) t = 4$$

Solve for t

$$t=1$$

$$t=4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5-t by t.

$$5t-t^{2}=4$$

Subtract 4 from both sides.

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

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.

$$-t^{2}+5t-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square 5.

$$t=\frac{-5±\sqrt{25-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-5±\sqrt{25+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$t=\frac{-5±\sqrt{25-16}}{2\left(-1\right)}$$

Add 25 to -16.

$$t=\frac{-5±\sqrt{9}}{2\left(-1\right)}$$

Take the square root of 9.

$$t=\frac{-5±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-5±3}{-2}$$

Now solve the equation t=\frac{-5±3}{-2} when ± is plus. Add -5 to 3.

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

Divide -2 by -2.

$$t=1$$

Now solve the equation t=\frac{-5±3}{-2} when ± is minus. Subtract 3 from -5.

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

Divide -8 by -2.

$$t=4$$

The equation is now solved.

$$t=1 t=4$$

Steps for Completing the Square

Use the distributive property to multiply 5-t by t.

$$5t-t^{2}=4$$

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.

$$-t^{2}+5t=4$$

Divide both sides by -1.

$$\frac{-t^{2}+5t}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\frac{5}{-1}t=\frac{4}{-1}$$

Divide 5 by -1.

$$t^{2}-5t=\frac{4}{-1}$$

Divide 4 by -1.

$$t^{2}-5t=-4$$

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-5t+\frac{25}{4}=-4+\frac{25}{4}$$

Add -4 to \frac{25}{4}.

$$t^{2}-5t+\frac{25}{4}=\frac{9}{4}$$

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

$$\left(t-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$t-\frac{5}{2}=\frac{3}{2} t-\frac{5}{2}=-\frac{3}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$t=4 t=1$$