Question

use the discriminant to determine the number of solutions to the quadratic equation −6z2−10z−3=0. What are the real solutions and complex solutions?

Answer 1

Explanation:

-6z²-10z-3=0

multiply by -1

6z²+10z+3=0

disc .=b²-4ac=10²-4×6×3=100-72=28≥0

also it is not a perfect square.

so roots are real,irrational and different.

`z=(-6 \pm√(28) )/(2 * 6) \\=(-6 \pm 2 √(7))/(12) \\=(-3 \pm√(7) )/(6)`