The problem is to solve the equation $y = b^2 + 10b = 1$. We need to find the value of $b$ that satisfies this equation.

1. Problem Description

The problem is to solve the equation $y = b^2 + 10b = 1$. We need to find the value of $b$ that satisfies this equation.

2. Solution Steps

First, rewrite the equation to equal zero by subtracting 1 from both sides:

$b^2 + 10b - 1 = 0$

This is a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = 10$, and $c = -1$. We can use the quadratic formula to solve for $b$:

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

Substitute the values of $a$, $b$, and $c$ into the quadratic formula:

$b = \frac{-10 \pm \sqrt{10^2 - 4(1)(-1)}}{2(1)}$

$b = \frac{-10 \pm \sqrt{100 + 4}}{2}$

$b = \frac{-10 \pm \sqrt{104}}{2}$

Simplify the square root:

$\sqrt{104} = \sqrt{4 \cdot 26} = 2\sqrt{26}$

Substitute this back into the equation:

$b = \frac{-10 \pm 2\sqrt{26}}{2}$

Divide both terms in the numerator by 2:

$b = -5 \pm \sqrt{26}$

Therefore, there are two possible solutions for $b$:

$b = -5 + \sqrt{26}$ and $b = -5 - \sqrt{26}$

3. Final Answer

The solutions are $b = -5 + \sqrt{26}$ and $b = -5 - \sqrt{26}$.