How do you solve #2x^2+4x-30 =0#?

2 Answers
Tazwar Sikder
Sep 5, 2016

#x = -5, 3#

Explanation:

We have: #2 x^(2) + 4 x - 30 = 0#

Let's apply the quadratic formula:

#=> x = (- 4 pm sqrt(4^(2) - 4 (2) (- 30))) / (2 (2))#

#=> x = (- 4 pm sqrt(16 + 240)) / (4)#

#=> x = (- 4 pm sqrt(256)) / (4)#

#=> x = (- 4 pm 16) / (4)#

#=> x = - 1 pm 4#

#=> x = - 5, 3#

Therefore, the solutions to the equation are #x = - 5# and #x = 3#.

EZ as pi
Sep 5, 2016

#x=-5 " or " x = 3#

Explanation:

All the terms in the equation are even, so we can divide by 2 immediately to make the numbers smaller.

#2x^2+ 4x -30 =0#

#x^2 +2x -15 = 0#

Try to factor before we use loner methods of the quadratic formula or completing the square.

Find factors of 15 which subtract (because of -15) to give 2.
The signs will be different (because of -15), there must be more positives. (because of +2)

5 and 3 will work! #5xx3 = 15 and 5-3 = 2#

#(x+5)(x-3) = 0#

Putting each factor equal to 0 gives the solutions:

#x=-5 " or " x = 3#

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