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

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Answer 1 · 2017-03-03T00:53:32

I guess you want to factor, you just have to apply the quadratic formula.

#x=(-b+-sqrt(b^2-4(ac)))/(2a)#

#x=(-2+-sqrt(2^2-4(1*-4)))/(2*1)#

#x=(-2+-sqrt(4-4(-4)))/(2)#

#x=(-2+-sqrt(4+16))/(2)#

#x=(-2+-sqrt(20))/(2)#

#x =(-2-√20)/2=-1-√ 5 = -3.236#

x =#(-2+√20)/2=-1+√ 5 = 1.236#