Answer:
2 nonreal solutions
Explanation:
given a quadratic equation in standard form
ax² + bx + c = 0 (a ≠ 0 )
then the nature of the roots are determined by the discriminant
b² - 4ac
• if b² - 4ac > 0 then 2 real and irrational solutions
• if b² - 4ac > 0 and a perfect square then 2 real and rational solutions
• if b² - 4ac = 0 then 2 real and equal solutions
• if b² - 4ac < 0 then no real solutions
5x² - 2x + 6 = 0 ← in standard form
with a = 5 , b = - 2 , c = 6
b² - 4ac
= (- 2)² - (4 × 5 × 6)
= 4 - 120
= - 116
since b² - 4ac < 0
then there are 2 nonreal solutions to the equation