Solve 2d^2+4=5d

$$2 d ^ { 2 } + 4 = 5 d$$

Solve for d

$$d=\frac{5+\sqrt{7}i}{4}\approx 1.25+0.661437828i$$

$$d=\frac{-\sqrt{7}i+5}{4}\approx 1.25-0.661437828i$$

Steps Using the Quadratic Formula

Subtract 5d from both sides.

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

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.

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

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

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

Square -5.

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

Multiply -4 times 2.

$$d=\frac{-\left(-5\right)±\sqrt{25-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$d=\frac{-\left(-5\right)±\sqrt{25-32}}{2\times 2}$$

Add 25 to -32.

$$d=\frac{-\left(-5\right)±\sqrt{-7}}{2\times 2}$$

Take the square root of -7.

$$d=\frac{-\left(-5\right)±\sqrt{7}i}{2\times 2}$$

The opposite of -5 is 5.

$$d=\frac{5±\sqrt{7}i}{2\times 2}$$

Multiply 2 times 2.

$$d=\frac{5±\sqrt{7}i}{4}$$

Now solve the equation d=\frac{5±\sqrt{7}i}{4} when ± is plus. Add 5 to i\sqrt{7}.

$$d=\frac{5+\sqrt{7}i}{4}$$

Now solve the equation d=\frac{5±\sqrt{7}i}{4} when ± is minus. Subtract i\sqrt{7} from 5.

$$d=\frac{-\sqrt{7}i+5}{4}$$

The equation is now solved.

$$d=\frac{5+\sqrt{7}i}{4} d=\frac{-\sqrt{7}i+5}{4}$$

Steps for Completing the Square

Subtract 5d from both sides.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 2.

$$\frac{2d^{2}-5d}{2}=-\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$d^{2}-\frac{5}{2}d=-\frac{4}{2}$$

Divide -4 by 2.

$$d^{2}-\frac{5}{2}d=-2$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$d^{2}-\frac{5}{2}d+\frac{25}{16}=-2+\frac{25}{16}$$

Add -2 to \frac{25}{16}.

$$d^{2}-\frac{5}{2}d+\frac{25}{16}=-\frac{7}{16}$$

Factor d^{2}-\frac{5}{2}d+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(d-\frac{5}{4}\right)^{2}=-\frac{7}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(d-\frac{5}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Simplify.

$$d-\frac{5}{4}=\frac{\sqrt{7}i}{4} d-\frac{5}{4}=-\frac{\sqrt{7}i}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$d=\frac{5+\sqrt{7}i}{4} d=\frac{-\sqrt{7}i+5}{4}$$