1: x=3, x=1

2: x= -5

3: There are 2 real solutions.

4: There are 2 real solutions.

5: There are no real solutions.

6. There is 1 real solution.

7.

8. x= -6, x = -2

9. x = -1/6, x=1

10.

Explanation:

1. The quadratic formula is

Substituting our known information we have:

2. Rewriting the quadratic in standard form we have x²+10x-25=0. Substituting this into the quadratic formula gives us:

3. The discriminant is b²-4ac. For this problem, that is 20²-4(-4)(25)=400--400=800. Since this is greater than 0, there are 2 real solutions.

4. The discriminant in this problem is 7²-4(2)(-15)=49--120=49+120=169. This is greater than 0, so there are 2 real solutions.

5. The discriminant in this problem is 1²-4(-2)(-28)=1-224=-223. Since this is less than 0, there are no real solutions.

6. If the discriminant of a quadratic is 0, then by definition there is 1 real solution.

7. Rewriting the quadratic we have 3x²-4x-2=0. Using the quadratic formula we have:

8. Factoring this trinomial we want factors of 12 that sum to 8. 6*2 = 12 and 6+2=8, so those are our factors. This gives us:

(x+6)(x+2)=0

Using the zero product property we know that either x+6=0 or x+2=0. Solving these equations we get x= -6 or x= -2.

9. Factoring this trinomial we want factors of 6(-1)=-6 that sum to -5. (-6)(1)=-6 and -6+1=-5, so this is how we "split up" the x term:

6x²-6x+1x-1=0

We group together the first two and the last two terms:

(6x²-6x)+(1x-1)=0

Factor the GCF out of each group. In the first group, that is 6x:

6x(x-1)+(1x-1)=0

In the second group, the GCF is 1:

6x(x-1)+1(x-1)=0

Both terms have a factor of (x-1), so we can factor it out:

(x-1)(6x+1)=0

Using the zero product property, we know either x-1=0 or 6x+1=0. Solving these equations we get x=1 or x=-1/6.

10. Substituting our information into the quadratic formula we get: