Solve 5(2x-1)^2+9=0

$$5 ( 2 x - 1 ) ^ { 2 } + 9 = 0$$

Solve for x (complex solution)

$$x=-\frac{3\sqrt{5}i}{10}+\frac{1}{2}\approx 0.5-0.670820393i$$

$$x=\frac{3\sqrt{5}i}{10}+\frac{1}{2}\approx 0.5+0.670820393i$$

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}.

$$5\left(4x^{2}-4x+1\right)+9=0$$

Use the distributive property to multiply 5 by 4x^{2}-4x+1.

$$20x^{2}-20x+5+9=0$$

Add 5 and 9 to get 14.

$$20x^{2}-20x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -20 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 20\times 14}}{2\times 20}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 20\times 14}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-20\right)±\sqrt{400-80\times 14}}{2\times 20}$$

Multiply -80 times 14.

$$x=\frac{-\left(-20\right)±\sqrt{400-1120}}{2\times 20}$$

Add 400 to -1120.

$$x=\frac{-\left(-20\right)±\sqrt{-720}}{2\times 20}$$

Take the square root of -720.

$$x=\frac{-\left(-20\right)±12\sqrt{5}i}{2\times 20}$$

The opposite of -20 is 20.

$$x=\frac{20±12\sqrt{5}i}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{20±12\sqrt{5}i}{40}$$

Now solve the equation x=\frac{20±12\sqrt{5}i}{40} when ± is plus. Add 20 to 12i\sqrt{5}.

$$x=\frac{20+12\sqrt{5}i}{40}$$

Divide 20+12i\sqrt{5} by 40.

$$x=\frac{3\sqrt{5}i}{10}+\frac{1}{2}$$

Now solve the equation x=\frac{20±12\sqrt{5}i}{40} when ± is minus. Subtract 12i\sqrt{5} from 20.

$$x=\frac{-12\sqrt{5}i+20}{40}$$

Divide 20-12i\sqrt{5} by 40.

$$x=-\frac{3\sqrt{5}i}{10}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{5}i}{10}+\frac{1}{2} x=-\frac{3\sqrt{5}i}{10}+\frac{1}{2}$$

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}.

$$5\left(4x^{2}-4x+1\right)+9=0$$

Use the distributive property to multiply 5 by 4x^{2}-4x+1.

$$20x^{2}-20x+5+9=0$$

Add 5 and 9 to get 14.

$$20x^{2}-20x+14=0$$

Subtract 14 from both sides. Anything subtracted from zero gives its negation.

$$20x^{2}-20x=-14$$

Divide both sides by 20.

$$\frac{20x^{2}-20x}{20}=-\frac{14}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\left(-\frac{20}{20}\right)x=-\frac{14}{20}$$

Divide -20 by 20.

$$x^{2}-x=-\frac{14}{20}$$

Reduce the fraction \frac{-14}{20} to lowest terms by extracting and canceling out 2.

$$x^{2}-x=-\frac{7}{10}$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{7}{10}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-\frac{7}{10}+\frac{1}{4}$$

Add -\frac{7}{10} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=-\frac{9}{20}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{9}{20}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{9}{20}}$$

Simplify.

$$x-\frac{1}{2}=\frac{3\sqrt{5}i}{10} x-\frac{1}{2}=-\frac{3\sqrt{5}i}{10}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{3\sqrt{5}i}{10}+\frac{1}{2} x=-\frac{3\sqrt{5}i}{10}+\frac{1}{2}$$