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Home >> Polynomials >> Quadratic Equation >> Finding roots of Quadratic Equation >> Finding roots of Quadratic Equation or solution of quadratic equation by quadratic formula
The formula to find the roots of a quadratic equation is known as the Quadratic Formula:
Note: if b^{2}  4ac > 0 then only we can find the roots of quadratic equation with this formula.
Let's understand how we get this formula
There are two methods in this:
Method 1 : Completing the Square
Consider the following quadratic formula:
ax^{2} + bx + c = 0
Note : a ≠ 0
Divide both sides by "a" and we get:
or we can also write it as
x^{2} +  b a  x  +  ⎧ ⎩  b 2a  ⎫ ^{2} ⎭  x    ⎧ ⎩  b 2a  ⎫ ^{2} ⎭  x +  c a  = 0 
Now, here see the first three variable i.e
x^{2}  +  b a  x  +  ⎧ ⎩  b 2a  ⎫ ^{2} ⎭ 
You will notice that here (a + b)^{2} formula applies and we get:
⎧ ⎩  x  +  b 2a  ⎫ ^{2} ⎭    ⎧ ⎩  b 2a  ⎫ ^{2} ⎭  +  ⎧ ⎩  c a  ⎫ ⎭  = 0 
solving  ⎧ ⎩  b 2a  ⎫^{2} ⎭  we get: 
⎧ ⎩  x  +  b 2a  ⎫ ^{2} ⎭    ⎧ ⎩  b ^{2} 4a  ⎫ ⎭  +  ⎧ ⎩  c a  ⎫ ⎭  = 0 
⎧ ⎩  x  +  b 2a  ⎫ ^{2} ⎭    ⎧ ⎩  b ^{2}  4ac 4a^{2}  ⎫ ⎭  = 0 
Move   ⎧ ⎩  b ^{2}  4ac 4a^{2}  ⎫ ⎭  to RHS and we get 
⎧ ⎩  x  +  b 2a  ⎫ ^{2} ⎭  =  ⎧ ⎩  b ^{2}  4ac 4a^{2}  ⎫ ⎭ 
Take square root of both sides and we get:
x  +  b 2a  =  ± √ ( b ^{2}  4ac ) 2a^{2} 
Move  ⎧ ⎩  b 2a  ⎫ ⎭  to RHS  we get: 
x =    ⎧ ⎩  b 2a  ⎫ ⎭  ±  √ ( b ^{2}  4ac ) 2a^{2} 
On joining like terms we get:
x =  b ± √ ( b ^{2}  4ac ) 2a^{2} 
So, the roots of given equation is :
x =  b + √ ( b ^{2}  4ac ) 2a^{2}  or  x =  b  √ ( b ^{2}  4ac ) 2a^{2} 
Method 2: It is a quite old and shorter method being used by Indian Mathematicians:
Consider the following quadratic formula:
ax^{2} + bx + c = 0
multiply both sides with 4a and we get:
4a^{2} x^{2} + 4abx + 4ac = 0
move 4ac on the RHS and we get:
4a^{2} x^{2} + 4abx =  4ac
add b^{2} on both sides and we get:
4a^{2} x^{2} + 4abx + b^{2} = b^{2}  4ac
You will notice that on LHS (a + b)^{2} formula applies and we get:
(2ax + b)^{2} = b^{2}  4ac
Taking square root on both sides and we get:
2ax + b  = ±  √ ( b ^{2}  4ac ) 
Subtract b from side and we get:
2ax  = b ±  √ ( b ^{2}  4ac ) 
Divide both sides by 2a and we get:
x =  b ± √ ( b ^{2}  4ac ) 2a 
Now let's apply the above explained formula and find solution for the following quadratic equation:
Example : 2x^{2} 5x + 3 = 0
Solution: Find the values of a, b and c in the given equation and we get:
a = 2
b = 5
c = 3
Put the above values in above calculated quadratic formula and we get
x =  (5) ± √ ( 5 ^{2}  4 x 2 x 3 ) 2 x 2 
Calculating square root and we get:

