Which value of b will cause the quadratic equation x^2+bx+5=0 to have two real number solution

Every value of b>4.47 and b<-4.47 will cause the quadratic equation to have two real number solution.

Step-by-step explanation:

We have the quadratic function , and we have to find the value of b.

A quadratic function is , a quadratic function usually has two real solutions. You can find that solutions using Bhaskara's Formula:

and are real solutions of the quadratic equation if and only if:

If the quadratic equation doesn't have real solutions.

If    the quadratic equation has only one solution.

Then in this case to have two real number solutions:

We have , where a=1, b, c=5

Then,

Adding 20 in both sides of the equation:

Which is the same as:

Then,

Then every value of b>4.47 and b<-4.47 will cause the quadratic equation to have two real number solution.

For example b=-5 or b=5.

If you replace with b=-5 in

Then the quadratic function has two real number solutions.

b i think

step-by-step explanation:

not really sure

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation: