Question

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

Answer 1

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

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