Solve left(4xright)^2+18x=22

$${ \left(4x \right) }^{ 2 } +18x=22$$

Solve for x

$$x=\frac{\sqrt{433}-9}{16}\approx 0.738040753$$

$$x=\frac{-\sqrt{433}-9}{16}\approx -1.863040753$$

Steps Using the Quadratic Formula

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}+18x=22$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+18x=22$$

Subtract 22 from both sides.

$$16x^{2}+18x-22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 18 for b, and -22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 16\left(-22\right)}}{2\times 16}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 16\left(-22\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-18±\sqrt{324-64\left(-22\right)}}{2\times 16}$$

Multiply -64 times -22.

$$x=\frac{-18±\sqrt{324+1408}}{2\times 16}$$

Add 324 to 1408.

$$x=\frac{-18±\sqrt{1732}}{2\times 16}$$

Take the square root of 1732.

$$x=\frac{-18±2\sqrt{433}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-18±2\sqrt{433}}{32}$$

Now solve the equation x=\frac{-18±2\sqrt{433}}{32} when ± is plus. Add -18 to 2\sqrt{433}.

$$x=\frac{2\sqrt{433}-18}{32}$$

Divide -18+2\sqrt{433} by 32.

$$x=\frac{\sqrt{433}-9}{16}$$

Now solve the equation x=\frac{-18±2\sqrt{433}}{32} when ± is minus. Subtract 2\sqrt{433} from -18.

$$x=\frac{-2\sqrt{433}-18}{32}$$

Divide -18-2\sqrt{433} by 32.

$$x=\frac{-\sqrt{433}-9}{16}$$

The equation is now solved.

$$x=\frac{\sqrt{433}-9}{16} x=\frac{-\sqrt{433}-9}{16}$$

Steps for Completing the Square

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}+18x=22$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+18x=22$$

Divide both sides by 16.

$$\frac{16x^{2}+18x}{16}=\frac{22}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{18}{16}x=\frac{22}{16}$$

Reduce the fraction \frac{18}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{9}{8}x=\frac{22}{16}$$

Reduce the fraction \frac{22}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{9}{8}x=\frac{11}{8}$$

Divide \frac{9}{8}, the coefficient of the x term, by 2 to get \frac{9}{16}. Then add the square of \frac{9}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{8}x+\left(\frac{9}{16}\right)^{2}=\frac{11}{8}+\left(\frac{9}{16}\right)^{2}$$

Square \frac{9}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{8}x+\frac{81}{256}=\frac{11}{8}+\frac{81}{256}$$

Add \frac{11}{8} to \frac{81}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{8}x+\frac{81}{256}=\frac{433}{256}$$

Factor x^{2}+\frac{9}{8}x+\frac{81}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{16}\right)^{2}=\frac{433}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{16}\right)^{2}}=\sqrt{\frac{433}{256}}$$

Simplify.

$$x+\frac{9}{16}=\frac{\sqrt{433}}{16} x+\frac{9}{16}=-\frac{\sqrt{433}}{16}$$

Subtract \frac{9}{16} from both sides of the equation.

$$x=\frac{\sqrt{433}-9}{16} x=\frac{-\sqrt{433}-9}{16}$$