Solve 4x^2-5=4

$$4 { x }^{ 2 } -5=4$$

Solve for x

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

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

Steps Using Difference of Squares

Subtract 4 from both sides.

$$4x^{2}-5-4=0$$

Subtract 4 from -5 to get -9.

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

Consider 4x^{2}-9. Rewrite 4x^{2}-9 as \left(2x\right)^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2x-3\right)\left(2x+3\right)=0$$

To find equation solutions, solve 2x-3=0 and 2x+3=0.

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

Steps by Finding Square Root

Add 5 to both sides.

$$4x^{2}=4+5$$

Add 4 and 5 to get 9.

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

Divide both sides by 4.

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

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$4x^{2}-5-4=0$$

Subtract 4 from -5 to get -9.

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

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}.

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

Square 0.

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

Multiply -4 times 4.

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

Multiply -16 times -9.

$$x=\frac{0±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$x=\frac{0±12}{2\times 4}$$

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.

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