How do you solve #5x ^ { 2} + 10x + 1= 3#?

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Answer 1 · 2017-11-29T02:29:51

See a solution process below:

First, subtract #color(red)(3)# from each side of the equation to put the equation in standard form:

#5x^2 + 10x + 1 - color(red)(3) = 3 - color(red)(3)#

#5x^2 + 10x - 2 = 0#

Now, 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)(5)# for #color(red)(a)#

#color(blue)(10)# for #color(blue)(b)#

#color(green)(-2)# for #color(green)(c)# gives:

#x = (-color(blue)(10) +- sqrt(color(blue)(10)^2 - (4 * color(red)(5) * color(green)(-2))))/(2 * color(red)(5))#

#x = (-color(blue)(10) +- sqrt(100 - (-40)))/10#

#x = (-color(blue)(10) +- sqrt(100 + 40))/10#

#x = (-color(blue)(10) +- sqrt(140))/10#

#x = (-color(blue)(10) +- sqrt(4 * 35))/10#

#x = (-color(blue)(10))/10 +- (sqrt(4)sqrt(35))/10#

#x = -1 +- (2sqrt(35))/10#

#x = -1 +- sqrt(35)/5#

#x = -1 - sqrt(35)/5# and #x = -1 + sqrt(35)/5#