How do you solve 2x^2 - 3x + 4 = 0 using the quadratic formula?

3 Answers
t0hierry
Jan 20, 2017

x = 3 pm sqrt(7)/(2sqrt(2))

Explanation:

f(x) = (xsqrt(2) -3/sqrt(2))^2 + 4 - 9/4 = 0
if xsqrt(2) -3 /sqrt(2) = pm sqrt(7)/4
or x = 3 pm sqrt(7)/(2sqrt(2))

George C.
Jan 20, 2017

x = 3/4+-sqrt(23)/4i

Explanation:

Given:

2x^2-3x+4 = 0

Note that this is of the form:

ax^2+bx+c = 0

with a=2, b=-3 and c=4

The discriminant Delta of this quadratic is given by the formula:

Delta = b^2-4ac = (-3)^2-4(2)(4) = 9-32 = -23

Since Delta < 0 we can tell that this quadratic has no Real zeros - only Complex ones.

We can still use the quadratic formula to find them:

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

color(white)(x) = (-b+-sqrt(Delta))/(2a)

color(white)(x) = (3+-sqrt(-23))/4

color(white)(x) = (3+-sqrt(23)i)/4

color(white)(x) = 3/4+-sqrt(23)/4i

Jim H
Jan 20, 2017

2x^2-3x+4 = 0

The solutions to ax^2+bx+c=0 are given by

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

In this case, we get

x = (-(-3)+-sqrt((-3)^3-4(2)(4)))/(2(2))

= (3 +- sqrt(9-32))/4

= (3 +- sqrt (-23))/4

The solutions are imaginary

3/4 +- sqrt23/4 i

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