Solve Using the Quadratic Formula 3z^2+z-4=0

$3 z^{2} + z - 4 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 3$ , $b = 1$ , and $c = - 4$ into the quadratic formula and solve for $z$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (3 \cdot - 4)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

One to any power is one.

$z = \frac{- 1 \pm \sqrt{1 - 4 \cdot 3 \cdot - 4}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 4$ .

Multiply $- 4$ by $3$ .

$z = \frac{- 1 \pm \sqrt{1 - 12 \cdot - 4}}{2 \cdot 3}$

Multiply $- 12$ by $- 4$ .

$z = \frac{- 1 \pm \sqrt{1 + 48}}{2 \cdot 3}$

Add $1$ and $48$ .

$z = \frac{- 1 \pm \sqrt{49}}{2 \cdot 3}$

Rewrite $49$ as $7^{2}$ .

$z = \frac{- 1 \pm \sqrt{7^{2}}}{2 \cdot 3}$

Pull terms out from under the radical, assuming positive real numbers.

$z = \frac{- 1 \pm 7}{2 \cdot 3}$

Multiply $2$ by $3$ .

$z = \frac{- 1 \pm 7}{6}$

Step 4

The final answer is the combination of both solutions.

$z = 1, - \frac{4}{3}$