Solve x(2x+3)-4=2

$$x ( 2 x + 3 ) - 4 = 2$$

Solve for x

$$x = \frac{\sqrt{57} - 3}{4} \approx 1.137458609$$

$$x=\frac{-\sqrt{57}-3}{4}\approx -2.637458609$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by 2x+3.

$$2x^{2}+3x-4=2$$

Subtract 2 from both sides.

$$2x^{2}+3x-4-2=0$$

Subtract 2 from -4 to get -6.

$$2x^{2}+3x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-3±\sqrt{9+48}}{2\times 2}$$

Add 9 to 48.

$$x=\frac{-3±\sqrt{57}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±\sqrt{57}}{4}$$

Now solve the equation x=\frac{-3±\sqrt{57}}{4} when ± is plus. Add -3 to \sqrt{57}.

$$x=\frac{\sqrt{57}-3}{4}$$

Now solve the equation x=\frac{-3±\sqrt{57}}{4} when ± is minus. Subtract \sqrt{57} from -3.

$$x=\frac{-\sqrt{57}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{57}-3}{4} x=\frac{-\sqrt{57}-3}{4}$$

Steps for Completing the Square

Use the distributive property to multiply x by 2x+3.

$$2x^{2}+3x-4=2$$

Add 4 to both sides.

$$2x^{2}+3x=2+4$$

Add 2 and 4 to get 6.

$$2x^{2}+3x=6$$

Divide both sides by 2.

$$\frac{2x^{2}+3x}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{3}{2}x=\frac{6}{2}$$

Divide 6 by 2.

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

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=3+\frac{9}{16}$$

Add 3 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{57}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{57}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{57}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{57}}{4} x+\frac{3}{4}=-\frac{\sqrt{57}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{57}-3}{4} x=\frac{-\sqrt{57}-3}{4}$$