How do you solve 3z ^ { 2} + 4z - 7= 0?

Apr 4, 2018

$z = - \frac{7}{3} \mathmr{and} z = 1$

Explanation:

Using the quadratic equation

z_(1,2) = (-b +- sqrt(b^2 –4ac))/(2a)

you get

$z = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \times 3 \times \left(- 7\right)}}{2 \times 3}$

$z = \frac{- 4 \pm \sqrt{16 + 84}}{6}$

$z = \frac{- 4 \pm \sqrt{100}}{6}$

$z = \frac{- 4 \pm 10}{6}$

$z = - \frac{14}{6} \mathmr{and} z = \frac{6}{6}$

$z = - \frac{14}{6} \mathmr{and} z = 1$

$z = - \frac{7}{3} \mathmr{and} z = 1$

Apr 4, 2018

1, and $- \frac{7}{3}$

Explanation:

$3 {z}^{2} + 4 z - 7 = 0$
Use shortcut when a + b + c = 0
$z = 1$, and $z = \frac{c}{a} = - \frac{7}{3}$

Reminder of shortcut --> $y = a {x}^{2} + b x + c = 0$

1. When a + b + c = 0
$x 1 = 1$, and $x 2 = \frac{c}{a}$
Example: $y = 3 {x}^{2} + 4 x - 7 = 0$
$x 1 = 1$, and $x 2 = - \frac{7}{3}$

2. When a - b + c = 0.
x1 = - 1, and $x 2 = - \frac{c}{a}$
Example. $y = 5 {x}^{2} + 3 x - 2 x = 0$
$x 1 = - 1$, and $x 2 = - \frac{c}{a} = \frac{2}{5}$