How do you solve $x^2 + 3x - 5 = 0$ ?

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Answer 1 · 2015-11-01T13:58:09

x = $(-3 + sqrt29)/2$ or $x = (-3 - sqrt29)/2$

This equation does not easily factorise so using the general equation $y = ax^2 + bx +c$ and the formula

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

then if $y = x^2 +3x-5$

a =1, b = 3 and c = -5

$x =( -3 + sqrt(3^2 - (-20)))/2$ or $x = (-3 - sqrt(3^2 - (-20)))/2$

which comes to the given answer.
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How do you solve x^2 + 3x - 5 = 0x^2 + 3x - 5 = 0?