How do you solve $2x^2 + 4x +10 = 0$ using the quadratic formula?

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Answer 1 · 2016-05-11T06:00:27

The solutions for the equation are:
$color(blue)( x = -1 + 2i$

$color(blue)(x = -1 - 2i$

$2x^2 + 4x + 10 = 0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:

$a=2, b=4 , c=10$

The Discriminant is given by:

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

$= (4)^2-(4 * 2 * 10)$

$= 16 - 80 = - 64$

The solutions are normally found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

$x = ((-4)+-sqrt(-64))/(2*2) = (-4+-8i)/4$

$= (-4+-8i)/4 = (4 ( -1 +- 2i))/ 4$

$= (cancel4 ( -1 +- 2i))/cancel 4$

$= -1 +- 2i$

$x = -1 + 2i$

$x = -1 - 2i$

How do you solve 2x^2 + 4x +10 = 02x^2 + 4x +10 = 0 using the quadratic formula?