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.

$x=2$	$y=$	$2^2-7times2+10=$	$4+14+10$	$=28$
$x=1$	$y=$	$1^2-7times1+10=$	$1+7+10$	$=18$
$x=0$	$y=$	$0^2-7times0+10=$	$0+0+10$	$=10$
$x=-1$	$y=$	$(-1)^2-7times(-1)+10=$	$1-7+10$	$=4$
$x=-2$	$y=$	$(-2)^2-7times(-2)+10=$	$4-14+10$	$=0$
$x=-3$	$y=$	$(-2)^2-7times(-2)+10=$	$9-21+10$	$=-2$

Using the quadratic formula solver Example 5

Check the answer

Quick stetch example 5