How do you solve #2y ^ { 2} - 5y = 12#?

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Answer 1 · 2017-04-22T12:08:16

See the entire solution process below:

First, subtract #color(red)(12)# from each side of the equation to put the equation in standard quadratic form:

#2y^2 - 5y - color(red)(12) = 12 - color(red)(12)#

#2y^2 - 5y - 12 = 0#

Next, we can factor the left side of the equation as:

#(2y + 3)(y - 4) = 0

Now, we can solve each term on the left side of the equation for #0# to find the solutions to the problem:

Solution 1)

#2y + 3 = 0#

#2y + 3 - color(red)(3) = 0 - color(red)(3)#

#2y + 0 = -3#

#2y = -3#

#(2y)/color(red)(2) = -3/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = -3/2#

#y = -3/2#

Solution 2)

#y - 4 = 0#

#y - 4 + color(red)(4) = 0 + color(red)(4)#

#y - 0 = 4#

#y = 4#

The solution is: #y = -3/2# and #y = 4#