For which of the following values of c does 5x^2 + c = 10 have no real solutions

a. 1
b. 5
c. 9
d.12

Please help me out, I'm completely stuck! After writing the answer, please add an explanation to it so that I can understand :) Thank you so much <3 :)

5x^2 -0*x +c-10=0
for no real solution D = b^2 -4ac<0
here a = 5 b= 0 C = (c-10)

D= 0^2 -4(5)(c-10) < 0
-20c +200<0
20c>200
c> 10................(i)
Hence c = 12 ................Ans

To solve use the quadratic equation and the b^2 - 4ac (discriminant) part; that tells you if the roots are real or imaginary. For this equation there is no b term, so we just compute -4ac. if -4ac is greater than zero, the roots are imaginary.

For c = 1, the equation is 5x^2 - 9 = 0. -4ac = -4(5)(-9) = -180
For c = 5, the equation is 5x^2 - 5 = 0. -4ac = -4(5)(-5) = -100
For c = 9, the equation is 5x^2 - 1 = 0. -4ac = -4(5)(-1) = -20
For c = 12, the equation is 5x^5 + 2 = 0. -4ac = -4(5)(2) = 20

Since you're subtracting 4ac, the first three answers give positive numbers, thus real roots. The fourth one gives a negative result when subtracted from b^2 (0), thus an imaginary result.

PS - when you have just the a and c terms, if the c term is positive and the a term is also positive (which it's supposed to be in standard form) the roots are always imaginary.

Looks like it's D since when plugged in:

5x² + 12 = 10

x² = -2/5

x would have complex solutions since the square of a real number can never be negative