Which values of c will cause the quadratic equation –x2 + 3x + c = 0 to have no real number solutions? check all that apply.
–5
negative nine-halves
negative one-quarter
1
startfraction 9 over 4 endfraction

-5

negative nine-halves

Step-by-step explanation:

we know that

In the quadratic equation

If

then

The system has no real numbers solutions

we have

so

substitute

Verify each case

case 1) -5

For c=-5

substitute

------> is true

therefore

The value of c=-5 will cause the quadratic equation  to have no real number solutions

case 2) negative nine-halves

For c=-9/2

substitute

------> is true

therefore

The value of c=-9/2 will cause the quadratic equation  to have no real number solutions

case 3) negative one-quarter

For c=-1/4

substitute

------> is not true

therefore

The value of c=-1/4 will not cause the quadratic equation  to have no real number solutions

case 4) 1

For c=1

substitute

------> is not true

therefore

The value of c=1 will not cause the quadratic equation  to have no real number solutions

case 5) 9 Over 4

For c=9/4

substitute

------> is not true

therefore

The value of c=9/4 will not cause the quadratic equation  to have no real number solutions

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol