Solve 2x^2-5x-1=x(4x

$$2 x ^ { 2 } - 5 x - 1 = x ( 4 x$$

Solve for x

$$x=\frac{\sqrt{17}-5}{4}\approx -0.219223594$$

$$x=\frac{-\sqrt{17}-5}{4}\approx -2.280776406$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$2x^{2}-5x-1=x^{2}\times 4$$

Subtract x^{2}\times 4 from both sides.

$$2x^{2}-5x-1-x^{2}\times 4=0$$

Combine 2x^{2} and -x^{2}\times 4 to get -2x^{2}.

$$-2x^{2}-5x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\left(-1\right)}}{2\left(-2\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\left(-1\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8\left(-1\right)}}{2\left(-2\right)}$$

Multiply 8 times -1.

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

Add 25 to -8.

$$x=\frac{-\left(-5\right)±\sqrt{17}}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{17}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{5±\sqrt{17}}{-4}$$

Now solve the equation x=\frac{5±\sqrt{17}}{-4} when ± is plus. Add 5 to \sqrt{17}.

$$x=\frac{\sqrt{17}+5}{-4}$$

Divide 5+\sqrt{17} by -4.

$$x=\frac{-\sqrt{17}-5}{4}$$

Now solve the equation x=\frac{5±\sqrt{17}}{-4} when ± is minus. Subtract \sqrt{17} from 5.

$$x=\frac{5-\sqrt{17}}{-4}$$

Divide 5-\sqrt{17} by -4.

$$x=\frac{\sqrt{17}-5}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{17}-5}{4} x=\frac{\sqrt{17}-5}{4}$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$2x^{2}-5x-1=x^{2}\times 4$$

Subtract x^{2}\times 4 from both sides.

$$2x^{2}-5x-1-x^{2}\times 4=0$$

Combine 2x^{2} and -x^{2}\times 4 to get -2x^{2}.

$$-2x^{2}-5x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$-2x^{2}-5x=1$$

Divide both sides by -2.

$$\frac{-2x^{2}-5x}{-2}=\frac{1}{-2}$$

Dividing by -2 undoes the multiplication by -2.

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

Divide -5 by -2.

$$x^{2}+\frac{5}{2}x=\frac{1}{-2}$$

Divide 1 by -2.

$$x^{2}+\frac{5}{2}x=-\frac{1}{2}$$

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.

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

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

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{1}{2}+\frac{25}{16}$$

Add -\frac{1}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{17}{16}$$

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}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{\sqrt{17}}{4} x+\frac{5}{4}=-\frac{\sqrt{17}}{4}$$

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

$$x=\frac{\sqrt{17}-5}{4} x=\frac{-\sqrt{17}-5}{4}$$