Solve d^2-d/4-3/8=0

$$d ^ { 2 } - d / 4 - 3 / 8 = 0$$

Solve for d

$$d=-\frac{1}{2}=-0.5$$

$$d=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 8, the least common multiple of 4,8.

$$8d^{2}-2d-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8d^{2}+ad+bd-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=8\left(-3\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -2.

$$a=-6 b=4$$

Rewrite 8d^{2}-2d-3 as \left(8d^{2}-6d\right)+\left(4d-3\right).

$$\left(8d^{2}-6d\right)+\left(4d-3\right)$$

Factor out 2d in 8d^{2}-6d.

$$2d\left(4d-3\right)+4d-3$$

Factor out common term 4d-3 by using distributive property.

$$\left(4d-3\right)\left(2d+1\right)$$

To find equation solutions, solve 4d-3=0 and 2d+1=0.

$$d=\frac{3}{4} d=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 8, the least common multiple of 4,8.

$$8d^{2}-2d-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$d=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Square -2.

$$d=\frac{-\left(-2\right)±\sqrt{4-4\times 8\left(-3\right)}}{2\times 8}$$

Multiply -4 times 8.

$$d=\frac{-\left(-2\right)±\sqrt{4-32\left(-3\right)}}{2\times 8}$$

Multiply -32 times -3.

$$d=\frac{-\left(-2\right)±\sqrt{4+96}}{2\times 8}$$

Add 4 to 96.

$$d=\frac{-\left(-2\right)±\sqrt{100}}{2\times 8}$$

Take the square root of 100.

$$d=\frac{-\left(-2\right)±10}{2\times 8}$$

The opposite of -2 is 2.

$$d=\frac{2±10}{2\times 8}$$

Multiply 2 times 8.

$$d=\frac{2±10}{16}$$

Now solve the equation d=\frac{2±10}{16} when ± is plus. Add 2 to 10.

$$d=\frac{12}{16}$$

Reduce the fraction \frac{12}{16} to lowest terms by extracting and canceling out 4.

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

Now solve the equation d=\frac{2±10}{16} when ± is minus. Subtract 10 from 2.

$$d=-\frac{8}{16}$$

Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

$$d=-\frac{1}{2}$$

The equation is now solved.

$$d=\frac{3}{4} d=-\frac{1}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 8, the least common multiple of 4,8.

$$8d^{2}-2d-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$8d^{2}-2d=3$$

Divide both sides by 8.

$$\frac{8d^{2}-2d}{8}=\frac{3}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$d^{2}+\left(-\frac{2}{8}\right)d=\frac{3}{8}$$

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

$$d^{2}-\frac{1}{4}d=\frac{3}{8}$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$d^{2}-\frac{1}{4}d+\left(-\frac{1}{8}\right)^{2}=\frac{3}{8}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$d^{2}-\frac{1}{4}d+\frac{1}{64}=\frac{3}{8}+\frac{1}{64}$$

Add \frac{3}{8} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$d^{2}-\frac{1}{4}d+\frac{1}{64}=\frac{25}{64}$$

Factor d^{2}-\frac{1}{4}d+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(d-\frac{1}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(d-\frac{1}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$d-\frac{1}{8}=\frac{5}{8} d-\frac{1}{8}=-\frac{5}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$d=\frac{3}{4} d=-\frac{1}{2}$$