How do you solve $x^{2} + 5 x - 2 = 0$ $x^2+5x-2=0$ by completing the square?

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Answer 1 · 2015-05-09T04:26:05

$f (x) = x^{2} + 5 x - 2 = 0$ $f(x) = x^2 + 5x - 2 = 0$

$f (x) = x^{2} + 5 x + \frac{25}{4} - \frac{25}{4} - 2 = 0$ $f(x) = x^2 + 5x + 25/4 - 25/4 - 2 = 0$

f(x) = (x + 5/2)^2 - 33/4 = 0

$(x + \frac{5}{2}) = + \frac{s q r 33}{4} \to x = - \frac{5}{2} + \frac{s q r 33}{4}$ $(x + 5/2) = + (sqr33)/4 -> x = -5/2 + (sqr33)/4$

$x + \frac{5}{2} = - \frac{s q r 33}{4} \to x = - \frac{5}{2} - \frac{s q r 33}{4}$ $x + 5/2 = -(sqr33)/4 -> x = -5/2 - (sqr33)/4$

How do you solve x2+5x−2=0x^2+5x-2=0 by completing the square?