How do you solve #2x^2 +3x + 7 = 0 #?

Question

Eva Adamson · Accepted Answer

See a solution process below:

We can use the quadratic equation to solve this problem: The quadratic formula states: For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by: #x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))# Substituting: #color(red)(2)# for #color(red)(a)# #color(blue)(3)# for #color(blue)(b)# #color(green)(7)# for #color(green)(c)# gives: #x = (-color(blue)(3) +- sqrt(color(blue)(3)^2 - (4 * color(red)(2) * color(green)(7))))/(2 * color(red)(2))# #x = (-color(blue)(3) +- sqrt(9 - 56))/4# #x = (-color(blue)(3) +- sqrt(-47))/4#

Julia Adams · Answer

undefined You can solve the quadratic equation 2x^2 + 3x + 7 = 0 using the quadratic formula, which is given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where a, b, and c are the coefficients of the quadratic equation (ax^2 + bx + c = 0).

For the equation 2x^2 + 3x + 7 = 0:
a = 2, b = 3, and c = 7.

Plugging these values into the quadratic formula:

x = (-3 ± √(3^2 - 4 * 2 * 7)) / (2 * 2)
x = (-3 ± √(9 - 56)) / 4
x = (-3 ± √(-47)) / 4

Since the discriminant (b^2 - 4ac) is negative, the solutions will involve complex numbers:

x = (-3 ± √(-47)i) / 4

So, the solutions are:

x = (-3 + √(-47)i) / 4 and x = (-3 - √(-47)i) / 4