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.