Solve (4x+1)^2=8x+18

$$( 4 x + 1 ) ^ { 2 } = 8 x + 18$$

Solve for x

$$x = \frac{\sqrt{17}}{4} \approx 1.030776406$$

$$x = -\frac{\sqrt{17}}{4} \approx -1.030776406$$

Steps by Finding Square Root

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=8x+18$$

Subtract 8x from both sides.

$$16x^{2}+8x+1-8x=18$$

Combine 8x and -8x to get 0.

$$16x^{2}+1=18$$

Subtract 1 from both sides.

$$16x^{2}=18-1$$

Subtract 1 from 18 to get 17.

$$16x^{2}=17$$

Divide both sides by 16.

$$x^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{17}}{4} x=-\frac{\sqrt{17}}{4}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=8x+18$$

Subtract 8x from both sides.

$$16x^{2}+8x+1-8x=18$$

Combine 8x and -8x to get 0.

$$16x^{2}+1=18$$

Subtract 18 from both sides.

$$16x^{2}+1-18=0$$

Subtract 18 from 1 to get -17.

$$16x^{2}-17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 0 for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 16\left(-17\right)}}{2\times 16}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 16\left(-17\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{0±\sqrt{-64\left(-17\right)}}{2\times 16}$$

Multiply -64 times -17.

$$x=\frac{0±\sqrt{1088}}{2\times 16}$$

Take the square root of 1088.

$$x=\frac{0±8\sqrt{17}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{0±8\sqrt{17}}{32}$$

Now solve the equation x=\frac{0±8\sqrt{17}}{32} when ± is plus.

$$x=\frac{\sqrt{17}}{4}$$

Now solve the equation x=\frac{0±8\sqrt{17}}{32} when ± is minus.

$$x=-\frac{\sqrt{17}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}}{4} x=-\frac{\sqrt{17}}{4}$$