Solve (2x-3)*(x+11)-17=-9-2^3

$$( 2 x - 3 ) \cdot ( x + 11 ) - 17 = - 9 - 2 ^ { 3 }$$

Solve for x

$$x=-11$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-3 by x+11 and combine like terms.

$$2x^{2}+19x-33-17=-9-2^{3}$$

Subtract 17 from -33 to get -50.

$$2x^{2}+19x-50=-9-2^{3}$$

Calculate 2 to the power of 3 and get 8.

$$2x^{2}+19x-50=-9-8$$

Subtract 8 from -9 to get -17.

$$2x^{2}+19x-50=-17$$

Add 17 to both sides.

$$2x^{2}+19x-50+17=0$$

Add -50 and 17 to get -33.

$$2x^{2}+19x-33=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 19 for b, and -33 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-19±\sqrt{19^{2}-4\times 2\left(-33\right)}}{2\times 2}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 2\left(-33\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-19±\sqrt{361-8\left(-33\right)}}{2\times 2}$$

Multiply -8 times -33.

$$x=\frac{-19±\sqrt{361+264}}{2\times 2}$$

Add 361 to 264.

$$x=\frac{-19±\sqrt{625}}{2\times 2}$$

Take the square root of 625.

$$x=\frac{-19±25}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-19±25}{4}$$

Now solve the equation x=\frac{-19±25}{4} when ± is plus. Add -19 to 25.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-19±25}{4} when ± is minus. Subtract 25 from -19.

$$x=-\frac{44}{4}$$

Divide -44 by 4.

$$x=-11$$

The equation is now solved.

$$x=\frac{3}{2} x=-11$$

Steps for Completing the Square

Use the distributive property to multiply 2x-3 by x+11 and combine like terms.

$$2x^{2}+19x-33-17=-9-2^{3}$$

Subtract 17 from -33 to get -50.

$$2x^{2}+19x-50=-9-2^{3}$$

Calculate 2 to the power of 3 and get 8.

$$2x^{2}+19x-50=-9-8$$

Subtract 8 from -9 to get -17.

$$2x^{2}+19x-50=-17$$

Add 50 to both sides.

$$2x^{2}+19x=-17+50$$

Add -17 and 50 to get 33.

$$2x^{2}+19x=33$$

Divide both sides by 2.

$$\frac{2x^{2}+19x}{2}=\frac{33}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{19}{2}x=\frac{33}{2}$$

Divide \frac{19}{2}, the coefficient of the x term, by 2 to get \frac{19}{4}. Then add the square of \frac{19}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{19}{2}x+\left(\frac{19}{4}\right)^{2}=\frac{33}{2}+\left(\frac{19}{4}\right)^{2}$$

Square \frac{19}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{19}{2}x+\frac{361}{16}=\frac{33}{2}+\frac{361}{16}$$

Add \frac{33}{2} to \frac{361}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{19}{2}x+\frac{361}{16}=\frac{625}{16}$$

Factor x^{2}+\frac{19}{2}x+\frac{361}{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}.

$$\left(x+\frac{19}{4}\right)^{2}=\frac{625}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{4}\right)^{2}}=\sqrt{\frac{625}{16}}$$

Simplify.

$$x+\frac{19}{4}=\frac{25}{4} x+\frac{19}{4}=-\frac{25}{4}$$

Subtract \frac{19}{4} from both sides of the equation.

$$x=\frac{3}{2} x=-11$$