Solve x^2-5x/2-3=0

$$x ^ { 2 } - \frac { 5 x } { 2 } - 3 = 0$$

Solve for x

$$x = \frac{\sqrt{73} + 5}{4} \approx 3.386000936$$

$$x=\frac{5-\sqrt{73}}{4}\approx -0.886000936$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

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

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-\left(-5\right)±\sqrt{25+48}}{2\times 2}$$

Add 25 to 48.

$$x=\frac{-\left(-5\right)±\sqrt{73}}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{73}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±\sqrt{73}}{4}$$

Now solve the equation x=\frac{5±\sqrt{73}}{4} when ± is plus. Add 5 to \sqrt{73}.

$$x=\frac{\sqrt{73}+5}{4}$$

Now solve the equation x=\frac{5±\sqrt{73}}{4} when ± is minus. Subtract \sqrt{73} from 5.

$$x=\frac{5-\sqrt{73}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{73}+5}{4} x=\frac{5-\sqrt{73}}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Add 6 to both sides. Anything plus zero gives itself.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 6 by 2.

$$x^{2}-\frac{5}{2}x=3$$

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.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=3+\left(-\frac{5}{4}\right)^{2}$$

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

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=3+\frac{25}{16}$$

Add 3 to \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{73}{16}$$

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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{73}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{73}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{73}}{4} x-\frac{5}{4}=-\frac{\sqrt{73}}{4}$$

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

$$x=\frac{\sqrt{73}+5}{4} x=\frac{5-\sqrt{73}}{4}$$