Solve 4t^2-3t

$$4 t ^ { 2 } - 3 t$$

Factor

$$t\left(4t-3\right)$$

Solution Steps

Factor out t.

$$t\left(4t-3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$4t^{2}-3t=0$$

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.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\times 4}$$

Take the square root of \left(-3\right)^{2}.

$$t=\frac{-\left(-3\right)±3}{2\times 4}$$

The opposite of -3 is 3.

$$t=\frac{3±3}{2\times 4}$$

Multiply 2 times 4.

$$t=\frac{3±3}{8}$$

Now solve the equation t=\frac{3±3}{8} when ± is plus. Add 3 to 3.

$$t=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$t=\frac{3}{4}$$

Now solve the equation t=\frac{3±3}{8} when ± is minus. Subtract 3 from 3.

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

Divide 0 by 8.

$$t=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{4} for x_{1} and 0 for x_{2}.

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

Subtract \frac{3}{4} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4t^{2}-3t=4\times \frac{4t-3}{4}t$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4t^{2}-3t=\left(4t-3\right)t$$

Evaluate

$$t\left(4t-3\right)$$