The roots are 2+3i and 2-3i
Step-by-step explanation:
x = (-b ±√b²-4ac)/2a
⇒ x = (4±√16 - 4×1×13)/2
⇒ x = (4 ± √-36)2
⇒ x = 4 ±6i/2 = 2 + 3i, 2 - 3i
Answer:
No solution in R (if you consider complex numbers then the other answer is correct)
Step-by-step explanation:
quadratic formula:
x1 = [tex]\frac{-b + \sqrt{b^2 - 4ac} }{2a}[/tex]
x2 = [tex]\frac{-b - \sqrt{b^2 - 4ac} }{2a}[/tex]
calculate the discriminant b²-4ac first
(-4)² - 4*13 = 16 - 52 = - 36
square root of a negative number does not exist in R and a discriminant < 0 indicates no solutions in R
