Answer:
15 feet
Step-by-step explanation:
Let's go ahead and draw a sketch as seen below;
We can go ahead and solve for x using the Pythagorean theorem as seen below;
![\begin{gathered} 25^2=x^2+(x-5)^2 \\ 625=x^2+x^2-10x+25 \\ 625=2x^2-10x+25 \\ 2x^2-10x+25-625=0 \\ 2x^2-10x-600=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6a74qelrzvr4ymlecq6b364n3bfsvknwev.png)
Recall that a quadratic equation in standard form is given as;
![ax^2+bx+c=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/mvkhuzwnjhb4epaf7jjcoq2vi4zdi4350m.png)
Comparing both equations, we can see that a = 2, b = -10, and c = -600
We'll go ahead and use the quadratic formula to solve for x as seen below;
![x=(-b\pm√(b^2-4ac))/(2a)](https://img.qammunity.org/2023/formulas/mathematics/college/jr19ixi2zltkocy82qhxfiop5lyv4hzbkm.png)
![\begin{gathered} x=(-(-10)\pm√((-10)^2-4(2)(-600)))/(2(2)) \\ x=(10\pm√(4900))/(4) \\ x=(10+√(4900))/(4),(10-√(4900))/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/t6bb8kcg1afqjxyct6colubvn6329sg6av.png)
![\begin{gathered} x=20,-15 \\ \therefore x=20\text{ or }x=-15 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/50gkf0kcv0awlu9u769zusb9iax2hzqte0.png)
Since we're solving for distance, we'll go with the positive value of x which is 20. So x = 20 ft.
So the distance between the base of the ladder and the wall will be 15 ft (x - 5 = 20 - 5 = 15)