Solve 16+x=5x^2

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

Solve for x

$$x = \frac{\sqrt{321} + 1}{10} \approx 1.891647287$$

$$x=\frac{1-\sqrt{321}}{10}\approx -1.691647287$$

Steps Using the Quadratic Formula

Subtract 5x^{2} from both sides.

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

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.

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

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

$$x=\frac{-1±\sqrt{1^{2}-4\left(-5\right)\times 16}}{2\left(-5\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-5\right)\times 16}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-1±\sqrt{1+20\times 16}}{2\left(-5\right)}$$

Multiply 20 times 16.

$$x=\frac{-1±\sqrt{1+320}}{2\left(-5\right)}$$

Add 1 to 320.

$$x=\frac{-1±\sqrt{321}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-1±\sqrt{321}}{-10}$$

Now solve the equation x=\frac{-1±\sqrt{321}}{-10} when ± is plus. Add -1 to \sqrt{321}.

$$x=\frac{\sqrt{321}-1}{-10}$$

Divide -1+\sqrt{321} by -10.

$$x=\frac{1-\sqrt{321}}{10}$$

Now solve the equation x=\frac{-1±\sqrt{321}}{-10} when ± is minus. Subtract \sqrt{321} from -1.

$$x=\frac{-\sqrt{321}-1}{-10}$$

Divide -1-\sqrt{321} by -10.

$$x=\frac{\sqrt{321}+1}{10}$$

The equation is now solved.

$$x=\frac{1-\sqrt{321}}{10} x=\frac{\sqrt{321}+1}{10}$$

Steps for Completing the Square

Subtract 5x^{2} from both sides.

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

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -5.

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

Dividing by -5 undoes the multiplication by -5.

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

Divide 1 by -5.

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

Divide -16 by -5.

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

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{16}{5}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{16}{5}+\frac{1}{100}$$

Add \frac{16}{5} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{321}{100}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{321}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{321}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{\sqrt{321}}{10} x-\frac{1}{10}=-\frac{\sqrt{321}}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$x=\frac{\sqrt{321}+1}{10} x=\frac{1-\sqrt{321}}{10}$$