How do you solve #x^ { 2} + 3x - 6.75= 0#?

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Answer 1 · 2017-12-29T02:21:49

#x=3/2, -9/2#

Probably the easiest way to go would be using the quadratic formula:

# x = (-b \pm sqrt(b^2-4ac)) / (2a) #

We have #a=1, b=-3, c=-6.75#

# x = (-3 \pm sqrt((3)^2-4(1)(-6.75))) / (2(1)) #

# x = (-3 \pm sqrt(9+27)) / 2 #

# x = (-3 \pm sqrt(36)) / 2 #

# x = (-3 \pm 6) / 2 #

#x=3/2, -9/2#

Which we can see here on the graph:

graph{x^2+3x-6.75}