# How do you solve -6z^2 + 7z + 3 = 0 using the quadratic formula?

Oct 13, 2015

$z = - 0.3333 \mathmr{and} z = 1.5$

#### Explanation:

${x}^{2} + b x + c = 0$

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

b = constant with variable to the power 1
a = constant with variable to the power squared
c = constant.

$- 6 {z}^{2} + 7 z + 3 = 0$
In this case:

a = -6
b = 7
c= 3

put all these values

$\frac{- 7 \pm \sqrt{{\left(7\right)}^{2} - 4 \left(- 6\right) \left(3\right)}}{2 \left(- 6\right)}$
$\frac{- 7 \pm \sqrt{49 + 72}}{-} 12$
$\frac{- 7 \pm \sqrt{121}}{-} 12$
$\frac{- 7 \pm 11}{-} 12$
$\frac{- 7 + 11}{-} 12 , \frac{- 7 - 11}{-} 12$
$\frac{4}{-} 12 , - \frac{18}{-} 12$
$z = - 0.3333 \mathmr{and} z = 1.5$