How do you solve $x^2+5x-4=0$ using the quadratic formula?

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Answer 1 · 2016-03-30T16:33:46

$x= color(blue)( (-5+sqrt(41))/2$

$x= color(blue)( (-5-sqrt(41))/2$

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

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

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (5)^2-(4*1 * (-4))$

$= 25 + 16 = 41$

The solutions are found using the formula:

$x=(-b+-sqrtDelta)/(2*a)$

$x = ((-5)+-sqrt(41))/(2*1) = (-5+-sqrt(41))/2$

$x= color(blue)( (-5+sqrt(41))/2$

$x= color(blue)( (-5-sqrt(41))/2$

How do you solve x^2+5x-4=0x^2+5x-4=0 using the quadratic formula?