Solve (2)(x+3)^2+3(x+3)-4=0

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

Solve for x

$$x=\frac{\sqrt{41}-15}{4}\approx -2.149218941$$

$$x=\frac{-\sqrt{41}-15}{4}\approx -5.350781059$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2\left(x^{2}+6x+9\right)+3\left(x+3\right)-4=0$$

Use the distributive property to multiply 2 by x^{2}+6x+9.

$$2x^{2}+12x+18+3\left(x+3\right)-4=0$$

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

$$2x^{2}+12x+18+3x+9-4=0$$

Combine 12x and 3x to get 15x.

$$2x^{2}+15x+18+9-4=0$$

Add 18 and 9 to get 27.

$$2x^{2}+15x+27-4=0$$

Subtract 4 from 27 to get 23.

$$2x^{2}+15x+23=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 15 for b, and 23 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\times 2\times 23}}{2\times 2}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 2\times 23}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-15±\sqrt{225-8\times 23}}{2\times 2}$$

Multiply -8 times 23.

$$x=\frac{-15±\sqrt{225-184}}{2\times 2}$$

Add 225 to -184.

$$x=\frac{-15±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-15±\sqrt{41}}{4}$$

Now solve the equation x=\frac{-15±\sqrt{41}}{4} when ± is plus. Add -15 to \sqrt{41}.

$$x=\frac{\sqrt{41}-15}{4}$$

Now solve the equation x=\frac{-15±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from -15.

$$x=\frac{-\sqrt{41}-15}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{41}-15}{4} x=\frac{-\sqrt{41}-15}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2\left(x^{2}+6x+9\right)+3\left(x+3\right)-4=0$$

Use the distributive property to multiply 2 by x^{2}+6x+9.

$$2x^{2}+12x+18+3\left(x+3\right)-4=0$$

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

$$2x^{2}+12x+18+3x+9-4=0$$

Combine 12x and 3x to get 15x.

$$2x^{2}+15x+18+9-4=0$$

Add 18 and 9 to get 27.

$$2x^{2}+15x+27-4=0$$

Subtract 4 from 27 to get 23.

$$2x^{2}+15x+23=0$$

Subtract 23 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+15x=-23$$

Divide both sides by 2.

$$\frac{2x^{2}+15x}{2}=-\frac{23}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{15}{2}x=-\frac{23}{2}$$

Divide \frac{15}{2}, the coefficient of the x term, by 2 to get \frac{15}{4}. Then add the square of \frac{15}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{15}{2}x+\left(\frac{15}{4}\right)^{2}=-\frac{23}{2}+\left(\frac{15}{4}\right)^{2}$$

Square \frac{15}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=-\frac{23}{2}+\frac{225}{16}$$

Add -\frac{23}{2} to \frac{225}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=\frac{41}{16}$$

Factor x^{2}+\frac{15}{2}x+\frac{225}{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}.

$$\left(x+\frac{15}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{15}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

$$x+\frac{15}{4}=\frac{\sqrt{41}}{4} x+\frac{15}{4}=-\frac{\sqrt{41}}{4}$$

Subtract \frac{15}{4} from both sides of the equation.

$$x=\frac{\sqrt{41}-15}{4} x=\frac{-\sqrt{41}-15}{4}$$