Question

use the quadratic formula to solve. express your answer in simplest form.

\\6b^2 + 6b - 6 = b\\

Explanation:

Rearrange the equation into standard form

To use the quadratic formula, we must first rewrite the equation in the standard form \(ax^2 + bx + c = 0\).
Subtract \(b\) from both sides of the given equation:

$$ 6b^2 + 6b - 6 = b $$

$$ 6b^2 + 5b - 6 = 0 $$

Here, the coefficients are \(a = 6\), \(b = 5\), and \(c = -6\).

Apply the quadratic formula

The quadratic formula to solve for the variable \(b\) is:

$$ b = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Substitute the values \(a = 6\), \(b = 5\), and \(c = -6\) into the formula:

$$ b = \frac{-5 \pm \sqrt{5^2 - 4(6)(-6)}}{2(6)} $$

Simplify the radicand

Calculate the value inside the square root:

$$ 5^2 - 4(6)(-6) = 25 - (-144) = 25 + 144 = 169 $$

Now substitute this back into the expression:

$$ b = \frac{-5 \pm \sqrt{169}}{12} $$

Find the simplified solutions

Since \(169\) is a perfect square (\(13^2 = 169\)), we simplify the square root:

$$ b = \frac{-5 \pm 13}{12} $$

This gives two distinct solutions:

$$ b_1 = \frac{-5 + 13}{12} = \frac{8}{12} = \frac{2}{3} $$

$$ b_2 = \frac{-5 - 13}{12} = \frac{-18}{12} = -\frac{3}{2} $$

Answer:

Use the quadratic formula to solve. Express your answer in simplest form.

\(6b^2 + 6b - 6 = b\)

\(b =\) <blank>\(\frac{2}{3}, -\frac{3}{2}\)</blank>