Solve x=-16x^2+81x+5

$$x=-16 { x }^{ 2 } +81x+5$$

Solve for x

$$x=\frac{\sqrt{105}}{4}+\frac{5}{2}\approx 5.061737691$$

$$x=-\frac{\sqrt{105}}{4}+\frac{5}{2}\approx -0.061737691$$

Steps Using the Quadratic Formula

Add 16x^{2} to both sides.

$$x+16x^{2}=81x+5$$

Subtract 81x from both sides.

$$x+16x^{2}-81x=5$$

Combine x and -81x to get -80x.

$$-80x+16x^{2}=5$$

Subtract 5 from both sides.

$$-80x+16x^{2}-5=0$$

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.

$$16x^{2}-80x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -80 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 16\left(-5\right)}}{2\times 16}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 16\left(-5\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-80\right)±\sqrt{6400-64\left(-5\right)}}{2\times 16}$$

Multiply -64 times -5.

$$x=\frac{-\left(-80\right)±\sqrt{6400+320}}{2\times 16}$$

Add 6400 to 320.

$$x=\frac{-\left(-80\right)±\sqrt{6720}}{2\times 16}$$

Take the square root of 6720.

$$x=\frac{-\left(-80\right)±8\sqrt{105}}{2\times 16}$$

The opposite of -80 is 80.

$$x=\frac{80±8\sqrt{105}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{80±8\sqrt{105}}{32}$$

Now solve the equation x=\frac{80±8\sqrt{105}}{32} when ± is plus. Add 80 to 8\sqrt{105}.

$$x=\frac{8\sqrt{105}+80}{32}$$

Divide 80+8\sqrt{105} by 32.

$$x=\frac{\sqrt{105}}{4}+\frac{5}{2}$$

Now solve the equation x=\frac{80±8\sqrt{105}}{32} when ± is minus. Subtract 8\sqrt{105} from 80.

$$x=\frac{80-8\sqrt{105}}{32}$$

Divide 80-8\sqrt{105} by 32.

$$x=-\frac{\sqrt{105}}{4}+\frac{5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{105}}{4}+\frac{5}{2} x=-\frac{\sqrt{105}}{4}+\frac{5}{2}$$

Steps for Completing the Square

Add 16x^{2} to both sides.

$$x+16x^{2}=81x+5$$

Subtract 81x from both sides.

$$x+16x^{2}-81x=5$$

Combine x and -81x to get -80x.

$$-80x+16x^{2}=5$$

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.

$$16x^{2}-80x=5$$

Divide both sides by 16.

$$\frac{16x^{2}-80x}{16}=\frac{5}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{80}{16}\right)x=\frac{5}{16}$$

Divide -80 by 16.

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

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{5}{16}+\frac{25}{4}$$

Add \frac{5}{16} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=\frac{105}{16}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{105}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{105}{16}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{105}}{4} x-\frac{5}{2}=-\frac{\sqrt{105}}{4}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{105}}{4}+\frac{5}{2} x=-\frac{\sqrt{105}}{4}+\frac{5}{2}$$