Using the quadratic formula solver Example 5

Solve           `x^2+7x+10=0`

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

Quick sketch

Therefore   `a=1` , `b=7`  &  `c=10`

`x=(-7+-sqrt(7^2-4times1times10))/(2times1)`

`x=(-7+-sqrt(49-40))/2`

`x=(-7+-sqrt9)/2`

`x=(-7+-3)/2`

`x=(-7+3)/2`   or  `x=(-7-3)/2`

`x=-4/2`   or  `x=-10/2`

`x=-2`   or  `x=-5`

 

Check the answer

`x^2+7x+10=0`

 

If `x=-2`              `(-2)^2+7times(-2)+10=0`

`4-14+10=0`  Which is correct

If `x=-5`              `(-5)^2+7times(-5)+10=0`

`25-35+10=0`   Which is correct

Answer:

The roots of `x^2+7x+10=0`   are  `x=-2`   and  `x=-5`

 

NOTE:

This example has also been used in completing the square examples and factoring quadratics (easy) to show that the roots `-2`  and `-5`  can be found using any of these methods.