Solve b^2+8b/3-9=0

$$b ^ { 2 } + \frac { 8 b } { 3 } - 9 = 0$$

Solve for b

$$b = \frac{\sqrt{97} - 4}{3} \approx 1.949619267$$

$$b=\frac{-\sqrt{97}-4}{3}\approx -4.616285934$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$3b^{2}+8b-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 8 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-8±\sqrt{8^{2}-4\times 3\left(-27\right)}}{2\times 3}$$

Square 8.

$$b=\frac{-8±\sqrt{64-4\times 3\left(-27\right)}}{2\times 3}$$

Multiply -4 times 3.

$$b=\frac{-8±\sqrt{64-12\left(-27\right)}}{2\times 3}$$

Multiply -12 times -27.

$$b=\frac{-8±\sqrt{64+324}}{2\times 3}$$

Add 64 to 324.

$$b=\frac{-8±\sqrt{388}}{2\times 3}$$

Take the square root of 388.

$$b=\frac{-8±2\sqrt{97}}{2\times 3}$$

Multiply 2 times 3.

$$b=\frac{-8±2\sqrt{97}}{6}$$

Now solve the equation b=\frac{-8±2\sqrt{97}}{6} when ± is plus. Add -8 to 2\sqrt{97}.

$$b=\frac{2\sqrt{97}-8}{6}$$

Divide -8+2\sqrt{97} by 6.

$$b=\frac{\sqrt{97}-4}{3}$$

Now solve the equation b=\frac{-8±2\sqrt{97}}{6} when ± is minus. Subtract 2\sqrt{97} from -8.

$$b=\frac{-2\sqrt{97}-8}{6}$$

Divide -8-2\sqrt{97} by 6.

$$b=\frac{-\sqrt{97}-4}{3}$$

The equation is now solved.

$$b=\frac{\sqrt{97}-4}{3} b=\frac{-\sqrt{97}-4}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$3b^{2}+8b-27=0$$

Add 27 to both sides. Anything plus zero gives itself.

$$3b^{2}+8b=27$$

Divide both sides by 3.

$$\frac{3b^{2}+8b}{3}=\frac{27}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$b^{2}+\frac{8}{3}b=\frac{27}{3}$$

Divide 27 by 3.

$$b^{2}+\frac{8}{3}b=9$$

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}+\frac{8}{3}b+\left(\frac{4}{3}\right)^{2}=9+\left(\frac{4}{3}\right)^{2}$$

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

$$b^{2}+\frac{8}{3}b+\frac{16}{9}=9+\frac{16}{9}$$

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

$$b^{2}+\frac{8}{3}b+\frac{16}{9}=\frac{97}{9}$$

Factor b^{2}+\frac{8}{3}b+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b+\frac{4}{3}\right)^{2}=\frac{97}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+\frac{4}{3}\right)^{2}}=\sqrt{\frac{97}{9}}$$

Simplify.

$$b+\frac{4}{3}=\frac{\sqrt{97}}{3} b+\frac{4}{3}=-\frac{\sqrt{97}}{3}$$

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

$$b=\frac{\sqrt{97}-4}{3} b=\frac{-\sqrt{97}-4}{3}$$