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

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

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Answer 1 · 2015-11-01T13:58:09

x = $\frac{- 3 + \sqrt{29}}{2}$ $(-3 + sqrt29)/2$ or $x = \frac{- 3 - \sqrt{29}}{2}$ $x = (-3 - sqrt29)/2$

Explanation:

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

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

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

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

$x = \frac{- 3 + \sqrt{3^{2} - (- 20)}}{2}$ $x =( -3 + sqrt(3^2 - (-20)))/2$ or $x = \frac{- 3 - \sqrt{3^{2} - (- 20)}}{2}$ $x = (-3 - sqrt(3^2 - (-20)))/2$

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

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