Solve x^2+3x+21=32+6x

$$x ^ { 2 } + 3 x + 21 = 32 + 6 x$$

Solve for x

$$x = \frac{\sqrt{53} + 3}{2} \approx 5.140054945$$

$$x=\frac{3-\sqrt{53}}{2}\approx -2.140054945$$

Steps Using the Quadratic Formula

Subtract 32 from both sides.

$$x^{2}+3x+21-32=6x$$

Subtract 32 from 21 to get -11.

$$x^{2}+3x-11=6x$$

Subtract 6x from both sides.

$$x^{2}+3x-11-6x=0$$

Combine 3x and -6x to get -3x.

$$x^{2}-3x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-11\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-11\right)}}{2}$$

Multiply -4 times -11.

$$x=\frac{-\left(-3\right)±\sqrt{9+44}}{2}$$

Add 9 to 44.

$$x=\frac{-\left(-3\right)±\sqrt{53}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{53}}{2}$$

Now solve the equation x=\frac{3±\sqrt{53}}{2} when ± is plus. Add 3 to \sqrt{53}.

$$x=\frac{\sqrt{53}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{53}}{2} when ± is minus. Subtract \sqrt{53} from 3.

$$x=\frac{3-\sqrt{53}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{53}+3}{2} x=\frac{3-\sqrt{53}}{2}$$

Steps for Completing the Square

Subtract 6x from both sides.

$$x^{2}+3x+21-6x=32$$

Combine 3x and -6x to get -3x.

$$x^{2}-3x+21=32$$

Subtract 21 from both sides.

$$x^{2}-3x=32-21$$

Subtract 21 from 32 to get 11.

$$x^{2}-3x=11$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=11+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=11+\frac{9}{4}$$

Add 11 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{53}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{53}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{53}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{53}}{2} x-\frac{3}{2}=-\frac{\sqrt{53}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{53}+3}{2} x=\frac{3-\sqrt{53}}{2}$$