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

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Answer 1 · 2015-08-16T16:45:19

#x_(1,2) = (-1 +- sqrt(41))/4#

You can solve this quadratic equation by using the quadratic formula, which tells you that for any general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

the two roots of the equation take the form

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)#

So, start by adding #-5# to both sides of the equation to get

#2x^2 + x - 5 = color(red)(cancel(color(black)(5))) - color(red)(cancel(color(black)(5)))#

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

Notice that you have #a=2#, #b=1#, and #c=-5#. This means that the two solutions will be

#x_(1,2) = (-1 +- sqrt(1^2 - 4 * 2 * (-5)))/(2 * 2)#

#x_(1,2) = color(green)((-1 +- sqrt(41))/4)#

You can simplify this if you want to get

#x_1 = (-1 + sqrt(41))/4 ~= 1.35078#

and

#x_2 = (-1 - sqrt(41))/4 ~= -1.85078#