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

Answer 1

Isaiah Anthony

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 #

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Answer 2

Julia Adams

To solve the quadratic equation (2x^2 + 4x + 10 = 0) using the quadratic formula, first identify the coefficients: (a = 2), (b = 4), and (c = 10). Then, substitute these values into the quadratic formula: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}). After substitution, solve for (x) by computing the values of the expression under the square root, and then apply the plus-minus operation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.