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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?

Sagot :

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.

[tex]z=\frac{-6 \pm\sqrt{28} }{2 \times 6} \\=\frac{-6 \pm 2 \sqrt{7}}{12} \\=\frac{-3 \pm\sqrt{7} }{6}[/tex]

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