As we know, algebra is one of the major part of mathematics which solves different types of equations like the quadratic equation, binomial equation, polynomial, etc.,. Algebra uses variables, say letters, to get the unknown value. For example, y + 5 = 1, where y is unknown. 

Some of the algebra formulas for different types of equations are given below:

1. a$^2$ - b$^2$ = (a + b)(a - b)
2. (a + b)$^2$ = a$^2$ + 2ab + b$^2$
3. (a - b)$^2$ = a$^2$ - 2ab + b$^2$
4. (a$^m$)$^n$ = amn = (a$^n$)$^m$
5. Quadratic equation: x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
6. Binomial Theorem:(a+b)$^{n}$ = a$^{n}$b$^{0}$ + na$^{n-1}$b$^{1}$ + $\frac{n(n-1)}{2!}$ a$^{n-2}$b$^{2}$ + ..... + a$^{0}$n$^{n}$

Based on Algebra formulas, some of the solved problems are given below:

Question 1: Solve: (2x+5)(2x-5)
Solution: 
Using the algebra formula,
a$^2$ - b$^2$ = (a + b)(a - b)

(2x+5)(2x-5) = (2x)$^2$ - (5)$^2$ 
(2x+5)(2x-5) = 4$^2$ - 25

Question 2: Find the roots of the equation: 2$x^2$ + 7$x$ - 15 = 0.
Solution:
Given equation: 2$x^2$ + 7$x$ - 15 = 0
where, a=2, b=7 & c=-15.

Using the quadratic equation,
x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

x = $\frac{-7\pm \sqrt{7^{2}-4(2)(-15)}}{2(2)}$

x = $\frac{-7\pm \sqrt{169}}{4}$

x = $\frac{-7\pm 13}{4}$

Therefore, x$_{1}$ = $\frac{-7 + 13}{4}$ = $\frac{3}{4}$ and

x$_{2}$ = $\frac{-7 - 13}{4}$ = -5