How do you solve #2x^2 + 7x = 5#?

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Answer 1 · 2017-01-31T02:17:32

#x=(-7+sqrt89)/4;# #x=(-7-sqrt89)/4#

Solve #2x^2+7x=5#

Subtract #5# from both sides.

#2x^2+7x-5=0#

This is a quadratic equation in the form #ax^2+7x-5#, where #a=2#, #b=7#, #c=-5#.

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

Plug in the values from the quadratic formula.

#x=(-7+-sqrt(7^2-4*2*-5))/(2*2)#

#x=(-7+-sqrt(49+40))/4#

#x=(-7+-sqrt(89))/(4)#

#89# is a prime number so it cannot be factored further.

#x=(-7+sqrt89)/4;# #x=(-7-sqrt89)/4#