Solve (2x+3/4)^2-9/16+5=0

$$( 2 x + \frac { 3 } { 4 } ) ^ { 2 } - \frac { 9 } { 16 } + 5 = 0$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{71}i-3}{8}\approx -0.375-1.053268722i$$

$$x=\frac{-3+\sqrt{71}i}{8}\approx -0.375+1.053268722i$$

Steps by Finding Square Root

Subtracting \frac{71}{16} from itself leaves 0.

$$\left(2x+\frac{3}{4}\right)^{2}=-\frac{71}{16}$$

Take the square root of both sides of the equation.

$$2x+\frac{3}{4}=\frac{\sqrt{71}i}{4} 2x+\frac{3}{4}=-\frac{\sqrt{71}i}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$2x+\frac{3}{4}-\frac{3}{4}=\frac{\sqrt{71}i}{4}-\frac{3}{4} 2x+\frac{3}{4}-\frac{3}{4}=-\frac{\sqrt{71}i}{4}-\frac{3}{4}$$

Subtracting \frac{3}{4} from itself leaves 0.

$$2x=\frac{\sqrt{71}i}{4}-\frac{3}{4} 2x=-\frac{\sqrt{71}i}{4}-\frac{3}{4}$$

Subtract \frac{3}{4} from \frac{i\sqrt{71}}{4}.

$$2x=\frac{-3+\sqrt{71}i}{4}$$

Subtract \frac{3}{4} from -\frac{i\sqrt{71}}{4}.

$$2x=\frac{-\sqrt{71}i-3}{4}$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{-3+\sqrt{71}i}{2\times 4} \frac{2x}{2}=\frac{-\sqrt{71}i-3}{2\times 4}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{-3+\sqrt{71}i}{2\times 4} x=\frac{-\sqrt{71}i-3}{2\times 4}$$

Divide \frac{i\sqrt{71}-3}{4} by 2.

$$x=\frac{-3+\sqrt{71}i}{8}$$

Divide \frac{-i\sqrt{71}-3}{4} by 2.

$$x=\frac{-\sqrt{71}i-3}{8}$$

The equation is now solved.

$$x=\frac{-3+\sqrt{71}i}{8} x=\frac{-\sqrt{71}i-3}{8}$$