Solve 16x^2+1=5x

$$16 x ^ { 2 } + 1 = 5 x$$

Solve for x (complex solution)

$$x=\frac{5+\sqrt{39}i}{32}\approx 0.15625+0.195156187i$$

$$x=\frac{-\sqrt{39}i+5}{32}\approx 0.15625-0.195156187i$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

$$16x^{2}+1-5x=0$$

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.

$$16x^{2}-5x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -5 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 16}}{2\times 16}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 16}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-5\right)±\sqrt{25-64}}{2\times 16}$$

Add 25 to -64.

$$x=\frac{-\left(-5\right)±\sqrt{-39}}{2\times 16}$$

Take the square root of -39.

$$x=\frac{-\left(-5\right)±\sqrt{39}i}{2\times 16}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{39}i}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{5±\sqrt{39}i}{32}$$

Now solve the equation x=\frac{5±\sqrt{39}i}{32} when ± is plus. Add 5 to i\sqrt{39}.

$$x=\frac{5+\sqrt{39}i}{32}$$

Now solve the equation x=\frac{5±\sqrt{39}i}{32} when ± is minus. Subtract i\sqrt{39} from 5.

$$x=\frac{-\sqrt{39}i+5}{32}$$

The equation is now solved.

$$x=\frac{5+\sqrt{39}i}{32} x=\frac{-\sqrt{39}i+5}{32}$$

Steps for Completing the Square

Subtract 5x from both sides.

$$16x^{2}+1-5x=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$16x^{2}-5x=-1$$

Divide both sides by 16.

$$\frac{16x^{2}-5x}{16}=-\frac{1}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}-\frac{5}{16}x=-\frac{1}{16}$$

Divide -\frac{5}{16}, the coefficient of the x term, by 2 to get -\frac{5}{32}. Then add the square of -\frac{5}{32} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{16}x+\left(-\frac{5}{32}\right)^{2}=-\frac{1}{16}+\left(-\frac{5}{32}\right)^{2}$$

Square -\frac{5}{32} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{16}x+\frac{25}{1024}=-\frac{1}{16}+\frac{25}{1024}$$

Add -\frac{1}{16} to \frac{25}{1024} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{16}x+\frac{25}{1024}=-\frac{39}{1024}$$

Factor x^{2}-\frac{5}{16}x+\frac{25}{1024}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{32}\right)^{2}=-\frac{39}{1024}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{32}\right)^{2}}=\sqrt{-\frac{39}{1024}}$$

Simplify.

$$x-\frac{5}{32}=\frac{\sqrt{39}i}{32} x-\frac{5}{32}=-\frac{\sqrt{39}i}{32}$$

Add \frac{5}{32} to both sides of the equation.

$$x=\frac{5+\sqrt{39}i}{32} x=\frac{-\sqrt{39}i+5}{32}$$