Solve (2x+3)(x-1)=15

$$(2x+3)(x-1)=15$$

Solve for x

$$x = \frac{\sqrt{145} - 1}{4} \approx 2.760398645$$

$$x=\frac{-\sqrt{145}-1}{4}\approx -3.260398645$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x+3 by x-1 and combine like terms.

$$2x^{2}+x-3=15$$

Subtract 15 from both sides.

$$2x^{2}+x-3-15=0$$

Subtract 15 from -3 to get -18.

$$2x^{2}+x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-18\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-18\right)}}{2\times 2}$$

Multiply -8 times -18.

$$x=\frac{-1±\sqrt{1+144}}{2\times 2}$$

Add 1 to 144.

$$x=\frac{-1±\sqrt{145}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±\sqrt{145}}{4}$$

Now solve the equation x=\frac{-1±\sqrt{145}}{4} when ± is plus. Add -1 to \sqrt{145}.

$$x=\frac{\sqrt{145}-1}{4}$$

Now solve the equation x=\frac{-1±\sqrt{145}}{4} when ± is minus. Subtract \sqrt{145} from -1.

$$x=\frac{-\sqrt{145}-1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{145}-1}{4} x=\frac{-\sqrt{145}-1}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 2x+3 by x-1 and combine like terms.

$$2x^{2}+x-3=15$$

Add 3 to both sides.

$$2x^{2}+x=15+3$$

Add 15 and 3 to get 18.

$$2x^{2}+x=18$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{1}{2}x=\frac{18}{2}$$

Divide 18 by 2.

$$x^{2}+\frac{1}{2}x=9$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=9+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=9+\frac{1}{16}$$

Add 9 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{145}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{145}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{145}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{145}}{4} x+\frac{1}{4}=-\frac{\sqrt{145}}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{145}-1}{4} x=\frac{-\sqrt{145}-1}{4}$$