Solve 2x(x-3)+(x+5)=0

$$2 x ( x - 3 ) + ( x + 5 ) = 0$$

Solve for x (complex solution)

$$x=\frac{5+\sqrt{15}i}{4}\approx 1.25+0.968245837i$$

$$x=\frac{-\sqrt{15}i+5}{4}\approx 1.25-0.968245837i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by x-3.

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

Combine -6x and x to get -5x.

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

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

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\times 5}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\times 5}}{2\times 2}$$

Multiply -8 times 5.

$$x=\frac{-\left(-5\right)±\sqrt{25-40}}{2\times 2}$$

Add 25 to -40.

$$x=\frac{-\left(-5\right)±\sqrt{-15}}{2\times 2}$$

Take the square root of -15.

$$x=\frac{-\left(-5\right)±\sqrt{15}i}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{15}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±\sqrt{15}i}{4}$$

Now solve the equation x=\frac{5±\sqrt{15}i}{4} when ± is plus. Add 5 to i\sqrt{15}.

$$x=\frac{5+\sqrt{15}i}{4}$$

Now solve the equation x=\frac{5±\sqrt{15}i}{4} when ± is minus. Subtract i\sqrt{15} from 5.

$$x=\frac{-\sqrt{15}i+5}{4}$$

The equation is now solved.

$$x=\frac{5+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+5}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 2x by x-3.

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

Combine -6x and x to get -5x.

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

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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

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

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{5}{2}+\frac{25}{16}$$

Add -\frac{5}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{15}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=-\frac{15}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{15}i}{4} x-\frac{5}{4}=-\frac{\sqrt{15}i}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{5+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+5}{4}$$