How do you solve $-x^2 - 3x + 5 = 0$ using the quadratic formula?

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Answer 1 · 2016-02-27T07:33:06

$x=( -3-sqrt(29))/(2) or x=( -3+sqrt(29))/(2)$

Given equation= $-x^2-3x+5=0$

Comparing with $ax^2+bx+c=0$

We get , $a= -1, b=-3, c=5$

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

$x=( 3+-sqrt(-3^2-4xx-1xx5))/(2xx-1)$

$x=( 3+-sqrt(9+20))/(-2)$

$x=( 3+-sqrt(29))/(-2)$

$x=( 3+sqrt(29))/(-2) or x=( 3-sqrt(29))/(-2)$

$x=( -3-sqrt(29))/(2) or x=( -3+sqrt(29))/(2)$

How do you solve -x^2 - 3x + 5 = 0-x^2 - 3x + 5 = 0 using the quadratic formula?