For each equation, determine the number of solutions and place on the appropriate field in the table. really need with this one

We know that
In a quadratic equation 
ax²+bx+c=0
the term (b² - 4ac) is called the "discriminant" because it can "discriminate" between the possible types of
so
(b² - 4ac) > 0 we get two real solutions
(b² - 4ac) =0  we get just ONE real solution
(b² - 4ac) < 0 we get complex solutions or no real solutions

part 1)
5x²+2=4x
5x²-4x+2=0
a=5
b=-4
c=2
(b² - 4ac)=(-4)²-4*(5)*(2)> 16-40> -24
so
(b² - 4ac) < 0> No Real Solutions

the answer Part 1) is
No Real Solutions

part 2) 
3(x+5)²=-2
3(x²+10x+25)=-2
3x²+30x+75=-2
3x²+30x+77=0
a=3
b=30
c=77
(b² - 4ac)=(30)²-4*(3)*(77)> 900-924 > -24
so
(b² - 4ac) < 0> No Real Solutions

the answer Part 2) is
No Real Solutions

part 3) 
4x²-16x=0
a=4
b=-16
c=0
(b² - 4ac)=(-16)²-4*(4)*(0)> 256 
so
(b² - 4ac) > 0> Two Real Solutions

the answer Part 3) is
Two Real Solutions

part 4) 
3x²+24x=-48
3x²+24x+48=0
a=3
b=24
c=48
(b² - 4ac)=(24)²-4*(3)*(48)> 0
so
(b² - 4ac) =0 > One Real Solution

the answer Part 4) is
One Real Solution

You can use a calculater and divide the minutes by the $$, hope that you