How do you solve $5 n^{2} - 10 n + 59 = 0$ $5n^2-10n+59=0$ ?

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Answer 1 · 2017-02-01T22:16:23

$x = 1 \pm \frac{3}{5} \sqrt{30} i$ $x=1+-3/5sqrt30i$

If the general form of the given equation is:

$a x^{2} + 2 n x + c$ $ax^2+2nx+c$ ,

then

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

Therefore:

$x = \frac{5 \pm \sqrt{5^{2} - 5 \cdot 59}}{5}$ $x=(5+-sqrt(5^2-5*59))/5$

$= \frac{5 \pm \sqrt{25 - 295}}{5}$ $=(5+-sqrt(25-295))/5$

The solutions are complex, since the number under the square root is negative, then

$x = \frac{5 \pm \sqrt{- 270}}{5} = 1 \pm \frac{3}{5} \sqrt{30} i$ $x=(5+-sqrt(-270))/5=1+-3/5sqrt30i$

How do you solve 5n2−10n+59=05n^2-10n+59=0?