What are the two solutions to the equation -0.2^2+12x+11=0? and how do you do them?
I am not going to do this for you. Use the quadratic equation.
Hint:
a = -.2
b = 12
c = 11
To find the solutions to the equation -0.2^2 + 12x + 11 = 0, we can use the quadratic formula. The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a).
In the given equation, a = -0.2^2 = -0.04, b = 12, and c = 11. Plugging these values into the quadratic formula, we get:
x = (-12 ± √(12^2 - 4 * (-0.04) * 11)) / (2 * (-0.04)).
Simplifying further:
x = (-12 ± √(144 + 0.44)) / (-0.08).
x = (-12 ± √144.44) / (-0.08).
Thus, the two solutions to the equation -0.2^2 + 12x + 11 = 0 can be calculated by evaluating the positive and negative results of the quadratic formula:
x₁ = (-12 + √144.44) / (-0.08),
x₂ = (-12 - √144.44) / (-0.08).
Evaluating these expressions using a calculator will give you the numerical values for x₁ and x₂.