Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.
-3y^2 + y = -6
rearranging things a bit, you have
3y^2 - y - 6 = 0
y = (1±√(1+72))/6
To solve the equation -3y^2 + y = -6 using the quadratic formula, we need to write it in the standard quadratic form, which is ax^2 + bx + c = 0.
In this case, we have -3y^2 + y = -6. To get it into standard form, let's move all terms to one side:
-3y^2 + y + 6 = 0
Now, we can identify the values of a, b, and c. In this equation, a = -3, b = 1, and c = 6. The quadratic formula is given by:
y = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
y = (-(1) ± √((1)^2 - 4(-3)(6))) / (2(-3))
Simplifying the equation further:
y = (-1 ± √(1 + 72)) / (-6)
y = (-1 ± √73) / (-6)
Now, let's solve for two possible values of y using the quadratic formula.
y1 = (-1 + √73) / (-6) ≈ 1.31
y2 = (-1 - √73) / (-6) ≈ -0.98
Therefore, the solutions to the equation -3y^2 + y = -6, rounded to the nearest hundredth, are y ≈ 1.31 and y ≈ -0.98.