Solve 144d^2-96d=-16

$$144 d ^ { 2 } - 96 d = - 16$$

Solve for d

$$d=\frac{1}{3}\approx 0.333333333$$

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.

$$144d^{2}-96d=-16$$

Add 16 to both sides of the equation.

$$144d^{2}-96d-\left(-16\right)=-16-\left(-16\right)$$

Subtracting -16 from itself leaves 0.

$$144d^{2}-96d-\left(-16\right)=0$$

Subtract -16 from 0.

$$144d^{2}-96d+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 144 for a, -96 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$d=\frac{-\left(-96\right)±\sqrt{\left(-96\right)^{2}-4\times 144\times 16}}{2\times 144}$$

Square -96.

$$d=\frac{-\left(-96\right)±\sqrt{9216-4\times 144\times 16}}{2\times 144}$$

Multiply -4 times 144.

$$d=\frac{-\left(-96\right)±\sqrt{9216-576\times 16}}{2\times 144}$$

Multiply -576 times 16.

$$d=\frac{-\left(-96\right)±\sqrt{9216-9216}}{2\times 144}$$

Add 9216 to -9216.

$$d=\frac{-\left(-96\right)±\sqrt{0}}{2\times 144}$$

Take the square root of 0.

$$d=-\frac{-96}{2\times 144}$$

The opposite of -96 is 96.

$$d=\frac{96}{2\times 144}$$

Multiply 2 times 144.

$$d=\frac{96}{288}$$

Reduce the fraction \frac{96}{288} to lowest terms by extracting and canceling out 96.

$$d=\frac{1}{3}$$

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.

$$144d^{2}-96d=-16$$

Divide both sides by 144.

$$\frac{144d^{2}-96d}{144}=-\frac{16}{144}$$

Dividing by 144 undoes the multiplication by 144.

$$d^{2}+\left(-\frac{96}{144}\right)d=-\frac{16}{144}$$

Reduce the fraction \frac{-96}{144} to lowest terms by extracting and canceling out 48.

$$d^{2}-\frac{2}{3}d=-\frac{16}{144}$$

Reduce the fraction \frac{-16}{144} to lowest terms by extracting and canceling out 16.

$$d^{2}-\frac{2}{3}d=-\frac{1}{9}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$d^{2}-\frac{2}{3}d+\left(-\frac{1}{3}\right)^{2}=-\frac{1}{9}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$d^{2}-\frac{2}{3}d+\frac{1}{9}=\frac{-1+1}{9}$$

Add -\frac{1}{9} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$d^{2}-\frac{2}{3}d+\frac{1}{9}=0$$

Factor d^{2}-\frac{2}{3}d+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(d-\frac{1}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(d-\frac{1}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$d-\frac{1}{3}=0 d-\frac{1}{3}=0$$

Add \frac{1}{3} to both sides of the equation.

$$d=\frac{1}{3} d=\frac{1}{3}$$

The equation is now solved. Solutions are the same.

$$d=\frac{1}{3}$$