Solve 9-4x^2=0

$$9 - 4 x ^ { 2 } = 0$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps by Finding Square Root

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}=-9$$

Divide both sides by -4.

$$x^{2}=\frac{-9}{-4}$$

Fraction \frac{-9}{-4} can be simplified to \frac{9}{4} by removing the negative sign from both the numerator and the denominator.

$$x^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$x=\frac{3}{2} x=-\frac{3}{2}$$

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.

$$-4x^{2}+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 0 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-4\right)\times 9}}{2\left(-4\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-4\right)\times 9}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16\times 9}}{2\left(-4\right)}$$

Multiply 16 times 9.

$$x=\frac{0±\sqrt{144}}{2\left(-4\right)}$$

Take the square root of 144.

$$x=\frac{0±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{0±12}{-8}$$

Now solve the equation x=\frac{0±12}{-8} when ± is plus. Reduce the fraction \frac{12}{-8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{3}{2}$$

Now solve the equation x=\frac{0±12}{-8} when ± is minus. Reduce the fraction \frac{-12}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{2}$$

The equation is now solved.

$$x=-\frac{3}{2} x=\frac{3}{2}$$