Let's define the following variable.
x = first integer
y = second integer
If the sum of their squares is 34, then we can form the equation below:

If the other integer, say x, is 1 less than twice the other (y), then we can form this second equation:

From these two equations, we can now solve for the values of x and y.
Here are the steps.
1. Replace the value of "x" in equation 1 using the value in equation 2.

2. Apply the exponent.

3. Rearrange the terms. Subtract 34 on both sides of the equation then, combine similar terms.

In step3, we are able to create a quadratic equation in a standard form that is ax² + bx + c wherein a = 5, b = -4, and c = -33.
To solve for the value of y, we can use the Quadratic formula.
![y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/gfo69urp8rw17yg122uzk6md9epfe1uf75.png)
Since we have already determined the values of a, b, and c, let's plug them into the formula above.
![y=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(5)(-33)}_{}}{2(5)}](https://img.qammunity.org/2023/formulas/mathematics/college/ieti41t21t27cocnxpxa671nz7vpwxfpi4.png)
Then, solve for y.
![\begin{gathered} y=\frac{4\pm\sqrt[]{16+660}}{10} \\ y=\frac{4\pm\sqrt[]{676}}{10} \\ y=(4\pm26)/(10) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fruijyau5s4274t0djvyv9b0fhjrik75ku.png)
Separate the plus and minus signs.

There are two possible values y but only one is an integer. So, we will use y = 3.
To solve for the other integer x, let's use the second equation plugin y = 3.

Therefore, the integers are 5 and 3.