How do you solve #2x^2 + 6x = 5# by using the quadratic formula?

1 Answer
Oct 20, 2015

Answer:

#x_(1,2) = (-6 +- sqrt(76))/4#

Explanation:

For a general form quadratic equation

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

you can calculate the roots by using the quadratic formula

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

So, the first thing to do here is rearrange your equation to match the general form. To do that, add #-5# to both sides of the equation

#2x^2 + 6x - 5 = color(red)(cancel(color(black)(5))) - color(red)(cancel(color(black)(5)))#

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

In your case, you have #a = 2#, #b = 6#, and #c = -5#. This means that the two solutions will take the form

#x_(1,2) = (-6 +- sqrt(6^2 - 4 * 2 * (-5)))/(2 * 2)#

#x_(1,2) = (-6 +- sqrt(76))/4#

SInce #76# is not a perfect square, the two solutions will be

#x_1 = (-6 - sqrt(76))/4" "# and #" "x_2 = (-6 +- sqrt(76))/4#

You an try out one to see if the calculations are correct

#2 * ((-6-sqrt(76))/4)^2 + 6 * ((-6 - sqrt(76))/4) - 5 = 0#

#2/16 * (36 + 2sqrt(76) + 76) + 3/2 * (-6 - sqrt(76)) -5 = 0#

#14 + color(red)(cancel(color(black)(3/2sqrt(76)))) - 9 - color(red)(cancel(color(black)(3/2sqrt(76)))) - 5 = 0#

#14 - 9 - 5 = 0 color(white)(x) color(green)(sqrt())#