Solve 2x(x-5)=3x-1

$$2 x ( x - 5 ) = 3 x - 1$$

Solve for x

$$x = \frac{\sqrt{161} + 13}{4} \approx 6.422144385$$

$$x=\frac{13-\sqrt{161}}{4}\approx 0.077855615$$

Steps Using the Quadratic Formula

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

$$2x^{2}-10x=3x-1$$

Subtract 3x from both sides.

$$2x^{2}-10x-3x=-1$$

Combine -10x and -3x to get -13x.

$$2x^{2}-13x=-1$$

Add 1 to both sides.

$$2x^{2}-13x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -13 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2}}{2\times 2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-13\right)±\sqrt{169-8}}{2\times 2}$$

Add 169 to -8.

$$x=\frac{-\left(-13\right)±\sqrt{161}}{2\times 2}$$

The opposite of -13 is 13.

$$x=\frac{13±\sqrt{161}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{13±\sqrt{161}}{4}$$

Now solve the equation x=\frac{13±\sqrt{161}}{4} when ± is plus. Add 13 to \sqrt{161}.

$$x=\frac{\sqrt{161}+13}{4}$$

Now solve the equation x=\frac{13±\sqrt{161}}{4} when ± is minus. Subtract \sqrt{161} from 13.

$$x=\frac{13-\sqrt{161}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{161}+13}{4} x=\frac{13-\sqrt{161}}{4}$$

Steps for Completing the Square

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

$$2x^{2}-10x=3x-1$$

Subtract 3x from both sides.

$$2x^{2}-10x-3x=-1$$

Combine -10x and -3x to get -13x.

$$2x^{2}-13x=-1$$

Divide both sides by 2.

$$\frac{2x^{2}-13x}{2}=-\frac{1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{13}{2}x=-\frac{1}{2}$$

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=-\frac{1}{2}+\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=-\frac{1}{2}+\frac{169}{16}$$

Add -\frac{1}{2} to \frac{169}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{161}{16}$$

Factor x^{2}-\frac{13}{2}x+\frac{169}{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}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{161}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{161}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{\sqrt{161}}{4} x-\frac{13}{4}=-\frac{\sqrt{161}}{4}$$

Add \frac{13}{4} to both sides of the equation.

$$x=\frac{\sqrt{161}+13}{4} x=\frac{13-\sqrt{161}}{4}$$