After substituting, expand $(x+1)^2$, combine like terms, and move everything to one side so the equation looks like $ax^2 + bx + c = 0$.

When you apply the quadratic formula, you will get two values for $x$. Use the context of the question: it specifically asks for the larger possible value of $x$.

In Desmos, type y = x + 1 on one line and x^2 + y^2 = 65 on another line. Desmos will graph a straight line and a circle.

Look for the points where the line and the circle intersect. Click or tap on each intersection point to see its coordinates $(x,y)$.

Compare the $x$-coordinates of the two intersection points. The larger of these $x$-values is the answer the question is asking for.

Use the second equation $y = x + 1$ to replace $y$ in the first equation.

Substitute $y = x + 1$:

Now expand $(x+1)^2$:

Combine like terms:

Subtract $65$ from both sides and simplify:

So $x$ must satisfy $x^2 + x - 32 = 0$.

You now have the quadratic equation

In the standard form $ax^2 + bx + c = 0$, we have $a = 1$, $b = 1$, and $c = -32$.

The quadratic formula says

Substitute $a = 1$, $b = 1$, and $c = -32$ into the formula (but do not simplify completely yet):

First simplify the expression under the square root (the discriminant):

So the square root part is $\sqrt{129}$. Its approximate value is

Using this approximate value in the quadratic formula, the two solutions for $x$ are approximately

• $x \approx \dfrac{-1 + 11.36}{2} \approx 5.18$
• $x \approx \dfrac{-1 - 11.36}{2} \approx -6.18$

The problem asks for the larger possible value of $x$, so we will take the positive solution and then write it in exact form.

From the quadratic formula with discriminant $129$, the exact solutions are

The larger solution corresponds to using the plus sign in the numerator (since $-1 + \sqrt{129} > -1 - \sqrt{129}$), so the larger possible value of $x$ is

This matches answer choice D.