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

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

Solve for x

$$x=\frac{\sqrt{85}-7}{6}\approx 0.369924076$$

$$x=\frac{-\sqrt{85}-7}{6}\approx -2.70325741$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5x-3 by x+1 and combine like terms.

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

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

$$5x^{2}+2x-3=2x^{2}-5x$$

Subtract 2x^{2} from both sides.

$$5x^{2}+2x-3-2x^{2}=-5x$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$3x^{2}+2x-3=-5x$$

Add 5x to both sides.

$$3x^{2}+2x-3+5x=0$$

Combine 2x and 5x to get 7x.

$$3x^{2}+7x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3\left(-3\right)}}{2\times 3}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 3\left(-3\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-7±\sqrt{49-12\left(-3\right)}}{2\times 3}$$

Multiply -12 times -3.

$$x=\frac{-7±\sqrt{49+36}}{2\times 3}$$

Add 49 to 36.

$$x=\frac{-7±\sqrt{85}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-7±\sqrt{85}}{6}$$

Now solve the equation x=\frac{-7±\sqrt{85}}{6} when ± is plus. Add -7 to \sqrt{85}.

$$x=\frac{\sqrt{85}-7}{6}$$

Now solve the equation x=\frac{-7±\sqrt{85}}{6} when ± is minus. Subtract \sqrt{85} from -7.

$$x=\frac{-\sqrt{85}-7}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{85}-7}{6} x=\frac{-\sqrt{85}-7}{6}$$

Steps for Completing the Square

Use the distributive property to multiply 5x-3 by x+1 and combine like terms.

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

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

$$5x^{2}+2x-3=2x^{2}-5x$$

Subtract 2x^{2} from both sides.

$$5x^{2}+2x-3-2x^{2}=-5x$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$3x^{2}+2x-3=-5x$$

Add 5x to both sides.

$$3x^{2}+2x-3+5x=0$$

Combine 2x and 5x to get 7x.

$$3x^{2}+7x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$3x^{2}+7x=3$$

Divide both sides by 3.

$$\frac{3x^{2}+7x}{3}=\frac{3}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{7}{3}x=\frac{3}{3}$$

Divide 3 by 3.

$$x^{2}+\frac{7}{3}x=1$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=1+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=1+\frac{49}{36}$$

Add 1 to \frac{49}{36}.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{85}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{85}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{85}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{\sqrt{85}}{6} x+\frac{7}{6}=-\frac{\sqrt{85}}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

$$x=\frac{\sqrt{85}-7}{6} x=\frac{-\sqrt{85}-7}{6}$$