Use the quadratic formula to solve the equation. If necessary round to the nearest hundredth.

Question

-2y^2 + 6y = -2

Answer 1

To solve the equation using the quadratic formula, we need to rearrange it into the form: ax^2 + bx + c = 0.

In this case, we have:
-2y^2 + 6y = -2

To rearrange it, we bring all terms to one side of the equation:
-2y^2 + 6y + 2 = 0

Now, we can identify the values of a, b, and c:
a = -2
b = 6
c = 2

Using the quadratic formula, which is x = (-b ± √(b^2 - 4ac)) / 2a, we can substitute in the values:

y = (-6 ± √(6^2 - 4(-2)(2))) / (2(-2))

Simplifying further:

y = (-6 ± √(36 + 16)) / (-4)
y = (-6 ± √52) / (-4)

Now, we can calculate the values by rounding to the nearest hundredth:

y ≈ (-6 + √52) / (-4) ≈ (-6 + 7.21) / (-4) ≈ 1.21 / (-4) ≈ -0.30

y ≈ (-6 - √52) / (-4) ≈ (-6 - 7.21) / (-4) ≈ -13.21 / (-4) ≈ 3.30

Therefore, the solutions to the equation -2y^2 + 6y = -2 (rounded to the nearest hundredth) are approximately -0.30 and 3.30.