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Solve using the quadratic 4g^(2)+4g+1=0 Write your answers as int

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Final answer:

The quadratic equation 4g^2+4g+1=0 has one real solution after applying the quadratic formula, which is g = -0.5.

Step-by-step explanation:

To solve the quadratic equation 4g^2+4g+1=0 using the quadratic formula, we identify the coefficients a, b, and c. In this equation, a=4, b=4, and c=1. The quadratic formula is given by:

x = (-b ± √(b^2-4ac)) / (2a)

Now, substituting the values of a, b, and c into the formula, we get:

g = (-(4) ± √((4)^2-4(4)(1))) / (2(4))

g = (-4 ± √(16-16)) / 8

g = (-4 ± √0) / 8

g = (-4 ± 0) / 8

g = -4 / 8

g = -0.5

Because the discriminant (b^2 - 4ac) is zero, there is one real solution to the equation, which is g = -0.5.

User Peadar Doyle
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