How do you solve #2x ^ { 2} + 5x + 6= 8#?

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Answer 1 · 2018-05-14T03:26:50

The two answers are: x = #(-5 + sqrt41) / 4# and #(-5 - sqrt41) / 4#

We need to start with #0# on the right, so simplify the equation:

#2x^2 + 5x + 6 = 8#
# 2x^2 + 5x + 6 - 8 = 0#
#2x^2 + 5x -2 = 0#

Now it is difficult to solve it by factorizing, so we need the following formula:

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

In our equation:
#a = 2#, #b = 5#, and #c = -2#.

Now, let's plug in the numbers.

#x = (-5 +- sqrt(5^2 - 4 * 2 * (-2))) / (2 * 2)#

#x = (-5 +- sqrt(25 + 16)) / 4#

#x = (-5 +- sqrt41)/4#