Please help me with this question algebra 1?

It's pretty simple actually.

A discriminant is b² - 4ac. You find the a,b,c values in the equation: ax²+bx+c

if b²-4ac is a positive number than there are 2 solutions

if b²-4ac is equal to 0 than there is 1 solution

if b²-4ac is a negative number than there are no solutions

Now for your questions:

#4.

b² - 4ac

(-3)²-4(3)(5) = 0

9-60=0

-51=0

***There are no solutions

#5.

b²-4ac

(6)²-4(-3)(-3)

36-36=0

0=0

***There is 1 solution

#6.

b²-4ac

(-5)²-4(1)(-10)

25+40=0

65=0

***There are 2 solutions

Hope i helped :)

The discriminant is b^2 - 4ac , where the quadratic is in the form ax^2+bx+c. If it is negative, there is no real solution. If it is 0, there is exactly one solution. If it is positive, there are two solutions.

4. a = 3, b =-3, c=5

(-3)^2 - 4*3*5

9 - 60 < 0, so no real solutions.

5. a=-3, b=6, c=-3

6^2 - 4*-3*-3

36-36 = 0, so exactly one real solution.

6. a=1, b=-5, c=-10

(-5)^2 - 4*1*-10

25+40 > 0, so two real solutions.