Solve 4-16x^2=0

$$4 - 16 x ^ { 2 } = 0$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=-\frac{1}{2}=-0.5$$

Steps by Finding Square Root

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

$$-16x^{2}=-4$$

Divide both sides by -16.

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

Reduce the fraction \frac{-4}{-16} to lowest terms by extracting and canceling out -4.

$$x^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$x=\frac{1}{2} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-16x^{2}+4=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-16\right)\times 4}}{2\left(-16\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-16\right)\times 4}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{0±\sqrt{64\times 4}}{2\left(-16\right)}$$

Multiply 64 times 4.

$$x=\frac{0±\sqrt{256}}{2\left(-16\right)}$$

Take the square root of 256.

$$x=\frac{0±16}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{0±16}{-32}$$

Now solve the equation x=\frac{0±16}{-32} when ± is plus. Reduce the fraction \frac{16}{-32} to lowest terms by extracting and canceling out 16.

$$x=-\frac{1}{2}$$

Now solve the equation x=\frac{0±16}{-32} when ± is minus. Reduce the fraction \frac{-16}{-32} to lowest terms by extracting and canceling out 16.

$$x=\frac{1}{2}$$

The equation is now solved.

$$x=-\frac{1}{2} x=\frac{1}{2}$$