How do you solve $x^{2} + 8 x - 5 = 0$ $x^2+8x-5=0$ ?

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How do you solve $x^{2} + 8 x - 5 = 0$ $x^2+8x-5=0$ ?

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Answer 1 · 2015-05-22T14:27:56

For a quadratic equation of the form $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$
the solutions are given by the quadratic formula
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x= (-b+-sqrt(b^2-4ac))/(2a)$

For $x^{2} + 8 x - 5 = 0$ $x^2+8x-5 = 0$
this becomes

$\frac{- 8 \pm \sqrt{64 + 20}}{2}$ $(-8+-sqrt(64+20))/2$

$= - 4 \pm \sqrt{21}$ $=-4+-sqrt(21)$

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How do you solve $x^{2} + 8 x - 5 = 0$ $x^2+8x-5=0$ ?

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How do you solve x2+8x−5=0x^2+8x-5=0?

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How do you solve $x^{2} + 8 x - 5 = 0$ $x^2+8x-5=0$ ?