How do you solve $y = x^{2} + 2 x - 4$ $y = x^2 + 2x - 4$ ?

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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.

$x = \frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$ $x=(-b+-sqrt(b^2-4(ac)))/(2a)$

$x = \frac{- 2 \pm \sqrt{2^{2} - 4 (1 \cdot - 4)}}{2 \cdot 1}$ $x=(-2+-sqrt(2^2-4(1*-4)))/(2*1)$

$x = \frac{- 2 \pm \sqrt{4 - 4 (- 4)}}{2}$ $x=(-2+-sqrt(4-4(-4)))/(2)$

$x = \frac{- 2 \pm \sqrt{4 + 16}}{2}$ $x=(-2+-sqrt(4+16))/(2)$

$x = \frac{- 2 \pm \sqrt{20}}{2}$ $x=(-2+-sqrt(20))/(2)$

$x = \frac{- 2 - \sqrt 20}{2} = - 1 - \sqrt 5 = - 3.236$ $x =(-2-√20)/2=-1-√ 5 = -3.236$

x = $\frac{- 2 + \sqrt 20}{2} = - 1 + \sqrt 5 = 1.236$ $(-2+√20)/2=-1+√ 5 = 1.236$

How do you solve y = x^2 + 2x - 4y=x2+2x−4y = x^2 + 2x - 4?