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

1 Answer
Oct 13, 2015

Answer:

#z = -0.3333 and z = 1.5 #

Explanation:

Standard Quadratic Equation:

#x^2 + bx + c = 0#

Quadratic formula:

#(-b+-sqrt(b^2 - 4ac)) / (2a)#

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

#-6z^2 + 7z + 3 = 0 #
In this case:

a = -6
b = 7
c= 3

put all these values

#(-7 +- sqrt((7)^2 - 4(-6)(3))) / (2(-6))#
#(-7 +- sqrt(49 + 72)) / -12 #
#(-7 +- sqrt(121)) / -12 #
#(-7 +- 11) / -12 #
#(-7 + 11) / -12 , (-7 - 11) / -12#
#4 / -12, -18 / -12 #
#z = -0.3333 and z = 1.5 #