How do you solve #3x^2 + 5x + 1 = 0#?

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Answer 1 · 2015-08-03T18:15:55

This equation has two real solutions:

#x_1=(-5-sqrt(13))/6#

#x_2=(-5+sqrt(13))/6#

To solve a quadratic equation you have to calculate #Delta#

#Delta=b^2-4ac#

#Delta=5^2-4*3*1=25-12=13#

Calculated value is positive, so the equation has 2 real solutions, which can be calculated using:

#x_(1,2)=(-b+- sqrt(Delta))/(4a)#