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

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?

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Tryphena 2 months 2021-07-28T08:33:19+00:00 1 Answers 2 views 0

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    2021-07-28T08:34:37+00:00

    Answer:

    Step-by-step 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=\frac{-6 \pm\sqrt{28} }{2 \times 6} \\=\frac{-6 \pm 2 \sqrt{7}}{12} \\=\frac{-3 \pm\sqrt{7} }{6}

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