Solve 4x^2+5=41

$$4 x ^ { 2 } + 5 = 41$$

Solve for x

$$x=3$$

$$x=-3$$

Steps Using Difference of Squares

Subtract 41 from both sides.

$$4x^{2}+5-41=0$$

Subtract 41 from 5 to get -36.

$$4x^{2}-36=0$$

Divide both sides by 4.

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

Consider x^{2}-9. Rewrite x^{2}-9 as x^{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).

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

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

$$x=3 x=-3$$

Steps by Finding Square Root

Subtract 5 from both sides.

$$4x^{2}=41-5$$

Subtract 5 from 41 to get 36.

$$4x^{2}=36$$

Divide both sides by 4.

$$x^{2}=\frac{36}{4}$$

Divide 36 by 4 to get 9.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

Steps Using the Quadratic Formula

Subtract 41 from both sides.

$$4x^{2}+5-41=0$$

Subtract 41 from 5 to get -36.

$$4x^{2}-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 4\left(-36\right)}}{2\times 4}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 4\left(-36\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16\left(-36\right)}}{2\times 4}$$

Multiply -16 times -36.

$$x=\frac{0±\sqrt{576}}{2\times 4}$$

Take the square root of 576.

$$x=\frac{0±24}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±24}{8}$$

Now solve the equation x=\frac{0±24}{8} when ± is plus. Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{0±24}{8} when ± is minus. Divide -24 by 8.

$$x=-3$$

The equation is now solved.

$$x=3 x=-3$$