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How do you solve #2x^2 - 3x + 4 = 0#?

Answer 1

This is a quadratic equation of the form: #ax^2+bx+c=0# Where: #a=2# #b=-3# #c=4# So you can use the Quadratic Formula: #x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)# #=(3+-sqrt(9-32))/4# The negative argument of the square root tells you that you will not get Real solutions (you'll get Complex Solutions). #x_(1,2)=(3+-sqrt(-1*23))/4=(3+-isqrt(23))/4# Where #i=sqrt(-1)#

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Answer 2

Kennedy Adams

To solve the quadratic equation 2x^2 - 3x + 4 = 0, you can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where: a = 2 b = -3 c = 4

Substitute the values of a, b, and c into the formula and solve for x.

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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.