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

1 Answer

Answer:

#x= (-3+sqrt(33))/4# or #x=(-3-sqrt(33))/4#

Explanation:

For a quadratic in the general standard form:
#color(white)("XXX")color(red)(a)x^2+color(blue)(b)x+color(green)(c)=0#
the quadratic formula for the solution(s) is
#color(white)("XXX")x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4(color(red)(a))(color(green)(c))))/(2(color(red)(a)))#

For the given equation: #color(red)(2)x^2+color(blue)(3)xcolor(green)(-3)#
the quadratic formula becomes:
#color(white)("XXX")x=(-color(blue)(3)+-sqrt(color(blue)(3)^2-4(color(red)(2))(color(green)(-3))))/(2(color(red)(2))#

#color(white)("XXX")x=(-color(blue)(3)+-sqrt(color(blue)9+24))/(2(color(red)(2))#

#color(white)("XXX")=(-3+-sqrt(33))/4#