Solve 4Tleft(3-Tright)=0

$$4T \left( 3-T \right) =0$$

Solve for T

$$T=3$$

$$T=0$$

Steps Using Factoring

Use the distributive property to multiply 4T by 3-T.

$$12T-4T^{2}=0$$

Factor out T.

$$T\left(12-4T\right)=0$$

To find equation solutions, solve T=0 and 12-4T=0.

$$T=0 T=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4T by 3-T.

$$12T-4T^{2}=0$$

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.

$$-4T^{2}+12T=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$T=\frac{-12±\sqrt{12^{2}}}{2\left(-4\right)}$$

Take the square root of 12^{2}.

$$T=\frac{-12±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$T=\frac{-12±12}{-8}$$

Now solve the equation T=\frac{-12±12}{-8} when ± is plus. Add -12 to 12.

$$T=\frac{0}{-8}$$

Divide 0 by -8.

$$T=0$$

Now solve the equation T=\frac{-12±12}{-8} when ± is minus. Subtract 12 from -12.

$$T=-\frac{24}{-8}$$

Divide -24 by -8.

$$T=3$$

The equation is now solved.

$$T=0 T=3$$

Steps for Completing the Square

Use the distributive property to multiply 4T by 3-T.

$$12T-4T^{2}=0$$

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.

$$-4T^{2}+12T=0$$

Divide both sides by -4.

$$\frac{-4T^{2}+12T}{-4}=\frac{0}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$T^{2}+\frac{12}{-4}T=\frac{0}{-4}$$

Divide 12 by -4.

$$T^{2}-3T=\frac{0}{-4}$$

Divide 0 by -4.

$$T^{2}-3T=0$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$T^{2}-3T+\left(-\frac{3}{2}\right)^{2}=\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$T^{2}-3T+\frac{9}{4}=\frac{9}{4}$$

Factor T^{2}-3T+\frac{9}{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}.

$$\left(T-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(T-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$T-\frac{3}{2}=\frac{3}{2} T-\frac{3}{2}=-\frac{3}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$T=3 T=0$$