Solve 3b^2+18b+17=b

$$3 b ^ { 2 } + 18 b + 17 = b$$

Solve for b

$$b=\frac{\sqrt{85}-17}{6}\approx -1.29674259$$

$$b=\frac{-\sqrt{85}-17}{6}\approx -4.369924076$$

Steps Using the Quadratic Formula

Subtract b from both sides.

$$3b^{2}+18b+17-b=0$$

Combine 18b and -b to get 17b.

$$3b^{2}+17b+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 17 for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-17±\sqrt{17^{2}-4\times 3\times 17}}{2\times 3}$$

Square 17.

$$b=\frac{-17±\sqrt{289-4\times 3\times 17}}{2\times 3}$$

Multiply -4 times 3.

$$b=\frac{-17±\sqrt{289-12\times 17}}{2\times 3}$$

Multiply -12 times 17.

$$b=\frac{-17±\sqrt{289-204}}{2\times 3}$$

Add 289 to -204.

$$b=\frac{-17±\sqrt{85}}{2\times 3}$$

Multiply 2 times 3.

$$b=\frac{-17±\sqrt{85}}{6}$$

Now solve the equation b=\frac{-17±\sqrt{85}}{6} when ± is plus. Add -17 to \sqrt{85}.

$$b=\frac{\sqrt{85}-17}{6}$$

Now solve the equation b=\frac{-17±\sqrt{85}}{6} when ± is minus. Subtract \sqrt{85} from -17.

$$b=\frac{-\sqrt{85}-17}{6}$$

The equation is now solved.

$$b=\frac{\sqrt{85}-17}{6} b=\frac{-\sqrt{85}-17}{6}$$

Steps for Completing the Square

Subtract b from both sides.

$$3b^{2}+18b+17-b=0$$

Combine 18b and -b to get 17b.

$$3b^{2}+17b+17=0$$

Subtract 17 from both sides. Anything subtracted from zero gives its negation.

$$3b^{2}+17b=-17$$

Divide both sides by 3.

$$\frac{3b^{2}+17b}{3}=-\frac{17}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$b^{2}+\frac{17}{3}b=-\frac{17}{3}$$

Divide \frac{17}{3}, the coefficient of the x term, by 2 to get \frac{17}{6}. Then add the square of \frac{17}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}+\frac{17}{3}b+\left(\frac{17}{6}\right)^{2}=-\frac{17}{3}+\left(\frac{17}{6}\right)^{2}$$

Square \frac{17}{6} by squaring both the numerator and the denominator of the fraction.

$$b^{2}+\frac{17}{3}b+\frac{289}{36}=-\frac{17}{3}+\frac{289}{36}$$

Add -\frac{17}{3} to \frac{289}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b^{2}+\frac{17}{3}b+\frac{289}{36}=\frac{85}{36}$$

Factor b^{2}+\frac{17}{3}b+\frac{289}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b+\frac{17}{6}\right)^{2}=\frac{85}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+\frac{17}{6}\right)^{2}}=\sqrt{\frac{85}{36}}$$

Simplify.

$$b+\frac{17}{6}=\frac{\sqrt{85}}{6} b+\frac{17}{6}=-\frac{\sqrt{85}}{6}$$

Subtract \frac{17}{6} from both sides of the equation.

$$b=\frac{\sqrt{85}-17}{6} b=\frac{-\sqrt{85}-17}{6}$$