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

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Answer 1 · 2017-10-23T16:59:51

See a solution process below:

We can use the quadratic equation to solve this problem:

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#