Solve left(2x-1right)^2+5=9x^2

$${ \left(2x-1 \right) }^{ 2 } +5=9 { x }^{ 2 }$$

Solve for x

$$x=\frac{\sqrt{34}-2}{5}\approx 0.766190379$$

$$x=\frac{-\sqrt{34}-2}{5}\approx -1.566190379$$

Steps Using the Quadratic Formula

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

$$4x^{2}-4x+1+5=9x^{2}$$

Add 1 and 5 to get 6.

$$4x^{2}-4x+6=9x^{2}$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+6-9x^{2}=0$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-4x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -4 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-5\right)\times 6}}{2\left(-5\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)\times 6}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-4\right)±\sqrt{16+20\times 6}}{2\left(-5\right)}$$

Multiply 20 times 6.

$$x=\frac{-\left(-4\right)±\sqrt{16+120}}{2\left(-5\right)}$$

Add 16 to 120.

$$x=\frac{-\left(-4\right)±\sqrt{136}}{2\left(-5\right)}$$

Take the square root of 136.

$$x=\frac{-\left(-4\right)±2\sqrt{34}}{2\left(-5\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{34}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{4±2\sqrt{34}}{-10}$$

Now solve the equation x=\frac{4±2\sqrt{34}}{-10} when ± is plus. Add 4 to 2\sqrt{34}.

$$x=\frac{2\sqrt{34}+4}{-10}$$

Divide 4+2\sqrt{34} by -10.

$$x=\frac{-\sqrt{34}-2}{5}$$

Now solve the equation x=\frac{4±2\sqrt{34}}{-10} when ± is minus. Subtract 2\sqrt{34} from 4.

$$x=\frac{4-2\sqrt{34}}{-10}$$

Divide 4-2\sqrt{34} by -10.

$$x=\frac{\sqrt{34}-2}{5}$$

The equation is now solved.

$$x=\frac{-\sqrt{34}-2}{5} x=\frac{\sqrt{34}-2}{5}$$

Steps for Completing the Square

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

$$4x^{2}-4x+1+5=9x^{2}$$

Add 1 and 5 to get 6.

$$4x^{2}-4x+6=9x^{2}$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+6-9x^{2}=0$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-4x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}-4x=-6$$

Divide both sides by -5.

$$\frac{-5x^{2}-4x}{-5}=-\frac{6}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{4}{-5}\right)x=-\frac{6}{-5}$$

Divide -4 by -5.

$$x^{2}+\frac{4}{5}x=-\frac{6}{-5}$$

Divide -6 by -5.

$$x^{2}+\frac{4}{5}x=\frac{6}{5}$$

Divide \frac{4}{5}, the coefficient of the x term, by 2 to get \frac{2}{5}. Then add the square of \frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=\frac{6}{5}+\left(\frac{2}{5}\right)^{2}$$

Square \frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{6}{5}+\frac{4}{25}$$

Add \frac{6}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{34}{25}$$

Factor x^{2}+\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{34}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{34}{25}}$$

Simplify.

$$x+\frac{2}{5}=\frac{\sqrt{34}}{5} x+\frac{2}{5}=-\frac{\sqrt{34}}{5}$$

Subtract \frac{2}{5} from both sides of the equation.

$$x=\frac{\sqrt{34}-2}{5} x=\frac{-\sqrt{34}-2}{5}$$