Solve 16x+0.1x^2+16=0

$$16 x + 0.1 x ^ { 2 } + 16 = 0$$

Solve for x

$$x=4\sqrt{390}-80\approx -1.006329367$$

$$x=-4\sqrt{390}-80\approx -158.993670633$$

Steps Using the Quadratic Formula

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.

$$0.1x^{2}+16x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.1 for a, 16 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 0.1\times 16}}{2\times 0.1}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 0.1\times 16}}{2\times 0.1}$$

Multiply -4 times 0.1.

$$x=\frac{-16±\sqrt{256-0.4\times 16}}{2\times 0.1}$$

Multiply -0.4 times 16.

$$x=\frac{-16±\sqrt{256-6.4}}{2\times 0.1}$$

Add 256 to -6.4.

$$x=\frac{-16±\sqrt{249.6}}{2\times 0.1}$$

Take the square root of 249.6.

$$x=\frac{-16±\frac{4\sqrt{390}}{5}}{2\times 0.1}$$

Multiply 2 times 0.1.

$$x=\frac{-16±\frac{4\sqrt{390}}{5}}{0.2}$$

Now solve the equation x=\frac{-16±\frac{4\sqrt{390}}{5}}{0.2} when ± is plus. Add -16 to \frac{4\sqrt{390}}{5}.

$$x=\frac{\frac{4\sqrt{390}}{5}-16}{0.2}$$

Divide -16+\frac{4\sqrt{390}}{5} by 0.2 by multiplying -16+\frac{4\sqrt{390}}{5} by the reciprocal of 0.2.

$$x=4\sqrt{390}-80$$

Now solve the equation x=\frac{-16±\frac{4\sqrt{390}}{5}}{0.2} when ± is minus. Subtract \frac{4\sqrt{390}}{5} from -16.

$$x=\frac{-\frac{4\sqrt{390}}{5}-16}{0.2}$$

Divide -16-\frac{4\sqrt{390}}{5} by 0.2 by multiplying -16-\frac{4\sqrt{390}}{5} by the reciprocal of 0.2.

$$x=-4\sqrt{390}-80$$

The equation is now solved.

$$x=4\sqrt{390}-80 x=-4\sqrt{390}-80$$

Steps for Completing the Square

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.

$$0.1x^{2}+16x+16=0$$

Subtract 16 from both sides of the equation.

$$0.1x^{2}+16x+16-16=-16$$

Subtracting 16 from itself leaves 0.

$$0.1x^{2}+16x=-16$$

Multiply both sides by 10.

$$\frac{0.1x^{2}+16x}{0.1}=-\frac{16}{0.1}$$

Dividing by 0.1 undoes the multiplication by 0.1.

$$x^{2}+\frac{16}{0.1}x=-\frac{16}{0.1}$$

Divide 16 by 0.1 by multiplying 16 by the reciprocal of 0.1.

$$x^{2}+160x=-\frac{16}{0.1}$$

Divide -16 by 0.1 by multiplying -16 by the reciprocal of 0.1.

$$x^{2}+160x=-160$$

Divide 160, the coefficient of the x term, by 2 to get 80. Then add the square of 80 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+160x+80^{2}=-160+80^{2}$$

Square 80.

$$x^{2}+160x+6400=-160+6400$$

Add -160 to 6400.

$$x^{2}+160x+6400=6240$$

Factor x^{2}+160x+6400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+80\right)^{2}=6240$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+80\right)^{2}}=\sqrt{6240}$$

Simplify.

$$x+80=4\sqrt{390} x+80=-4\sqrt{390}$$

Subtract 80 from both sides of the equation.

$$x=4\sqrt{390}-80 x=-4\sqrt{390}-80$$