How to solve y = -0.04x^2 + 1.3x + 56 using the quadratic formula?
Solve? You mean the zeroes?
y=0=-0.04x^2 + 1.3x + 56
x=(1.3 +-sqrt(1.3^2+3*.04*56)/(.08)
do the math.
Do you mean 4*0.4? And how did you get positive 0.4? That was the part that confused me.
To solve the equation y = -0.04x^2 + 1.3x + 56 using the quadratic formula, you need to determine the values of x that satisfy the equation. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here's how you can apply the quadratic formula to solve this equation:
1. Identify the values of a, b, and c in the equation y = -0.04x^2 + 1.3x + 56. In this case:
a = -0.04
b = 1.3
c = 56
2. Substitute these values into the quadratic formula:
x = (-1.3 ± √(1.3^2 - 4(-0.04)(56))) / (2(-0.04))
3. Simplify the equation inside the square root:
x = (-1.3 ± √(1.69 + 8.96)) / (-0.08)
4. Continue simplifying:
x = (-1.3 ± √10.65) / (-0.08)
5. Calculate the square root of 10.65:
√10.65 ≈ 3.266
6. Now, substitute this result back into the equation:
x = (-1.3 + 3.266) / (-0.08) --> x = 30.95
x = (-1.3 - 3.266) / (-0.08) --> x ≈ -49.43
So, the solutions to the equation y = -0.04x^2 + 1.3x + 56 are approximately x = 30.95 and x = -49.43.