How do you solve the quadratic with complex numbers given $- x^{2} + 4 x - 5 = 0$ $-x^2+4x-5=0$ ?

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Answer 1 · 2016-09-10T04:35:27

Use the quadratic formula. $x = 2 \pm i$ $x=2+-i$

$- x^{2} + 4 x - 5 = 0$ $-x^2+4x-5=0$

Use the quadratic formula

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

$a = - 1$ $a=-1$
$b = 4$ $b=4$
$c = - 5$ $c=-5$

$x = \frac{- 4 \pm \sqrt{4^{2} - (4 \cdot - 1 \cdot - 5)}}{2 \cdot - 1}$ $x=(-4+-sqrt(4^2-(4*-1*-5)))/(2*-1)$

$x = \frac{- 4 \pm \sqrt{16 - 20}}{- 2}$ $x=(-4+-sqrt(16-20))/-2$

$x = \frac{- 4 \pm \sqrt{- 4}}{- 2}$ $x=(-4+-sqrt(-4))/-2$

$x = \frac{- 4 \pm \sqrt{4} \sqrt{- 1}}{- 2}$ $x=(-4+-sqrt(4)sqrt(-1))/-2$

$\sqrt{- 1} = i$ $sqrt(-1) = i$

$x = \frac{- 4 \pm 2 i}{- 2}$ $x=(-4+-2i)/-2$

Divide both of the terms in the numerator by the denominator.

$x = \frac{- 4}{- 2} \pm \frac{2 i}{- 2}$ $x=(-4)/-2 +- (2i)/-2$

$x = 2 \pm i$ $x=2+-i$ which can also be written as $x = 2 + i$ $x=2+i$ and $x = 2 - i$ $x=2-i$

How do you solve the quadratic with complex numbers given -x^2+4x-5=0−x2+4x−5=0-x^2+4x-5=0?