How do you solve the quadratic #x^2+3x-28=0# using any method?

Question

8372 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-31T12:48:07

#x= 4, -7#

#x^2 +3 x -28 = 0#

#x^2 +7 x - 4 x -28= 0#

#x ( x+7 ) -4 (x + 7 ) = 0#

#( x+7) ( x-4 ) = 0#

Either # ( x+7) = 0, or ( x-4) = 0#

If # x+7 = 0#

#x = -7#

If #x-7 = 0#

#x= 4#

#x = 4 , -7#