Factor the trinomial `x^2 + 6x + 5`

`(x+5)(x+1)=0 `

Set each of the factors of the left-hand side of the equation equal to `0` .

`x+5=0`

`x+1=0 `

Since `5` does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting `5` from both sides.

`x=-5 `

`x+1=0 `

Set each of the factors of the left-hand side of the equation equal to `0` .

`x=-5 `

`x+1=0 `

Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting `1` from both sides.

`x=-5 `

`x=-1`

The complete solution is the set of the individual solutions.

`x=-5,-1`

**QUESTION:-**

** ****`x^2+6x+5=0`**

**``**

**SOLUTION:-**

**METHOD # 1:-**

Breaking down the quadratic equation;

`x^2+6x+5=0`

x^2 + 1x + 5x + 5 = 0

Take x and 5 common;

x(x+1) + 5(x+1) = 0

Therefore;

x + 1= 0 , x+5 =0

x = -1 , x = -5

Hence solved!

**METHOD # 2:-**

Solving for x by using quadratic formula;

x = {-b+- sqrt (b^2-4ac)}/(2a)`

x = {-(6)+- sqrt((6)^2 - 4(1)(5))} /(2*1)`

x = {-6+- sqrt (36-20)}/2

x = {-6 +- sqrt(16)} / 2

x = {-6 +- 4} /2

Take 2 common in numerator;

x = 2 (-3 +-2) /2

The 2 in numerator and the 2 in denominator are cancelled with each other;

x = -3 +-2

x = -3+2 , x = -3-2

x = -1 , x = -5

By both methods the answers we are getting are as follows;

x = -1

x = -5

Hence Solved.

x2 + 6x + 5 = 0

Since this is a quadratic equation we can solve it with the help of quadratic formula,

`x= [-b+-sqrt(b^2-4ac)] / (2a)`

Where

a= 1

b= 6

c= 5

Input the values in the formula,

`x= [-6+-sqrt(6^2-4*1*5)] / (2*1)`

`x= [-6+-sqrt(36-20)] / 2`

`x= [-6+-sqrt(16)] / 2`

`x= [-6+-4] / 2`

`x= [-6+4] / 2 => -2/2 => -1 `

`x= [-6-4] / 2 => -10/2 => -5`

**`Solution Set: {-1,-5}`**

` `

x^2+6x+5=0

what multiplies to get 5 and add to get 6

x^2+x+5x+5=0

(x^2+x)+(5x+5)=0

x(x+1)+5(x+1)=0

(X+5) (X+1)=0

x=-5, x=-1

`x^2+6x+5=0`

`x^2+1x+5x+5=0`

`(x^2+1x)(5x+5)=0`

`x(x+1)` `5(x+1)`

`(x+5) ` `(x+1)`

`x=-5` `x=-1`