Solve x*(2*x+5)=36

$$x \times (2 \times x+5)=36$$

Solve for x

$$x = \frac{\sqrt{313} - 5}{4} \approx 3.172951503$$

$$x=\frac{-\sqrt{313}-5}{4}\approx -5.672951503$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by 2x+5.

$$2x^{2}+5x=36$$

Subtract 36 from both sides.

$$2x^{2}+5x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\left(-36\right)}}{2\times 2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 2\left(-36\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-5±\sqrt{25-8\left(-36\right)}}{2\times 2}$$

Multiply -8 times -36.

$$x=\frac{-5±\sqrt{25+288}}{2\times 2}$$

Add 25 to 288.

$$x=\frac{-5±\sqrt{313}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-5±\sqrt{313}}{4}$$

Now solve the equation x=\frac{-5±\sqrt{313}}{4} when ± is plus. Add -5 to \sqrt{313}.

$$x=\frac{\sqrt{313}-5}{4}$$

Now solve the equation x=\frac{-5±\sqrt{313}}{4} when ± is minus. Subtract \sqrt{313} from -5.

$$x=\frac{-\sqrt{313}-5}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{313}-5}{4} x=\frac{-\sqrt{313}-5}{4}$$

Steps for Completing the Square

Use the distributive property to multiply x by 2x+5.

$$2x^{2}+5x=36$$

Divide both sides by 2.

$$\frac{2x^{2}+5x}{2}=\frac{36}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{5}{2}x=\frac{36}{2}$$

Divide 36 by 2.

$$x^{2}+\frac{5}{2}x=18$$

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.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=18+\left(\frac{5}{4}\right)^{2}$$

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

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=18+\frac{25}{16}$$

Add 18 to \frac{25}{16}.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{313}{16}$$

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}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{313}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{313}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{\sqrt{313}}{4} x+\frac{5}{4}=-\frac{\sqrt{313}}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$x=\frac{\sqrt{313}-5}{4} x=\frac{-\sqrt{313}-5}{4}$$