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please help solve this quadratic equation -9y²+12y-3=0​

Sagot :

Answer:

-y = 0.66 ± [tex]\sqrt{0.33}[/tex]

Step-by-step explanation:

Since the x-term coefficient is -9, we divide everything else by 9.

-y^2 + 1.33 - 0.33 = 0

Move the 0.33 to the right hand side of the equation.

-y^2 + 1.33 = -0.33

Take the 1.33, divide it by two, square it, and add it to both sides of the equation.

y^2 + 1.33 + 0.44 = -0.33 + 0.44

Complete the right side of the equation.

-y^2 + 1.33 + 0.44 = 0.11

Take 0.44, and subtract it by 0.11. In other words, move 0.44 to the right hand side of the equation.

-y^2 + 1.33 = 0.33

Take the square roots of both sides and solve.

[tex]\sqrt{(-y+0.66)^2[/tex]= [tex]\sqrt{0.33}[/tex]

-y + 0.66 = [tex]\sqrt{0.33}[/tex]

-y = 0.66 ± [tex]\sqrt{0.33}[/tex]

Really, really hope this helps!!