Solve 4x^2+65=81

$$4 x ^ { 2 } + 65 = 81$$

Solve for x

$$x=2$$

$$x=-2$$

Steps Using Difference of Squares

Subtract 81 from both sides.

$$4x^{2}+65-81=0$$

Subtract 81 from 65 to get -16.

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

Divide both sides by 4.

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

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

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

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

$$x=2 x=-2$$

Steps by Finding Square Root

Subtract 65 from both sides.

$$4x^{2}=81-65$$

Subtract 65 from 81 to get 16.

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

Divide both sides by 4.

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

Divide 16 by 4 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

Subtract 81 from both sides.

$$4x^{2}+65-81=0$$

Subtract 81 from 65 to get -16.

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

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

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

Square 0.

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

Multiply -4 times 4.

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

Multiply -16 times -16.

$$x=\frac{0±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

$$x=\frac{0±16}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±16}{8}$$

Now solve the equation x=\frac{0±16}{8} when ± is plus. Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{0±16}{8} when ± is minus. Divide -16 by 8.

$$x=-2$$

The equation is now solved.

$$x=2 x=-2$$