.2x^2 + .31x - 0.15 = 0
I don't see an easy factor here, Recommend the quadratic equation.
To solve the quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the given equation: 0.2x^2 + 0.31x - 0.15 = 0, we can identify the values of a, b, and c:
a = 0.2
b = 0.31
c = -0.15
Now we can substitute these values into the quadratic formula to find the solutions for x:
x = (-(0.31) ± √((0.31)^2 - 4 * 0.2 * (-0.15))) / (2 * 0.2)
Simplifying further:
x = (-0.31 ± √(0.0961 + 0.12)) / 0.4
x = (-0.31 ± √0.2161) / 0.4
Now we calculate the square root:
x = (-0.31 ± 0.4649) / 0.4
There are two possible solutions:
x = (-0.31 + 0.4649) / 0.4 ≈ 0.385
x = (-0.31 - 0.4649) / 0.4 ≈ -1.119
Therefore, the solutions to the equation 0.2x^2 + 0.31x - 0.15 = 0, using the quadratic formula, are approximately x = 0.385 and x = -1.119.