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larisa [96]
3 years ago
12

Solve the equation for all real solutions in simplest form. 3z^2+17z+16=0

Mathematics
1 answer:
Rudiy273 years ago
8 0

Answer:

x=\frac{-17+\sqrt{97}}{6}, x=\frac{-17-\sqrt{97}}{6}

Step-by-step explanation:

Use the quadratic formula to solve for all real solutions.

Quadratic formula: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Plug in and solve:

x=\frac{-17\pm\sqrt{17^2-4(3)(16)}}{2(3)}\\\\x=\frac{-17\pm\sqrt{289-4(3)(16)}}{2(3)}\\\\x=\frac{-17\pm\sqrt{289-12(16)}}{2(3)}\\\\x=\frac{-17\pm\sqrt{289-192}}{2(3)}\\\\x=\frac{-17\pm\sqrt{97}}{6}\\

Since that's the farthest we can simplify to, your answers would be the positive and negative versions of the expression:

x=\frac{-17+\sqrt{97}}{6}

and

x=\frac{-17-\sqrt{97}}{6}

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