# Solve the following quadrtatic equation. 5x²+2x+1=0 5x² + 2x + 1 = 0

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

a = 5, b = 2, c = 1

Substitute the variables into the quadratic formula and simplify.

[-2 +- sqrt(2^2 - 4 * 5 * 1)]...

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5x² + 2x + 1 = 0

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

a = 5, b = 2, c = 1

Substitute the variables into the quadratic formula and simplify.

[-2 +- sqrt(2^2 - 4 * 5 * 1)] / (2 * 5)

[-2 +- sqrt(4 - 20)] / 10

[-2 +- sqrt(-16)] / 10

Since you cannot square root a negative number, there are no real solutions.  However, if you are using imaginary numbers, proceed as follows:

(-2 +- 4i) / 10

-0.2 + 0.4i and -0.2 - 0.4i

Posted on We have to solve the equation: 5x^2+2x+1=0

Use the quadratic formula for the roots of the equation ax^2 + bx + c = 0, x1 = [-b + sqrt(b^2 - 4ac)]/2a and x2 = [-b - sqrt(b^2 - 4ac)]/2a

Here a = 5, b = 2 and c = 1

x1 = [-2 + sqrt(4 - 4*5)]/10 = -2/10 + [sqrt (-16)]/10 = -2/10 + [i*sqrt 16]/10 = -2/10 + 4*i/10 = -0.2 + 0.4*i

x2 = -0.2 - 0.4*i

The roots of the given equation are -0.2 + 0.4i and -0.2 - 0.4i

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