What are the solutions to the equation 3(1-y)^2+5(1-y)+2=0 ?

The solutions to the equation 3(1-y)^2+5(1-y)+2=0 are y=2,y= 5/3

Find the zeros of 3(1-y)^2+5(1-y)+2=0

The zeros of 3(1-y)^2+5(1-y)+2=0 are y=2,y= 5/3

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3 (1 - y)^{2} + 5 (1 - y) + 2 = 0

y = 2, y = \frac{5}{3}

Decimal Notation

y = 2, y = 1.66666 \dots

Solution steps

3 (1 - y)^{2} + 5 (1 - y) + 2 = 0

Expand

3 (1 - y)^{2} + 5 (1 - y) + 2 : 3 y^{2} - 11 y + 10

3 y^{2} - 11 y + 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 3 \cdot 10}{2 \cdot 3}

(- 11)^{2} - 4 \cdot 3 \cdot 10 = 1

y_{1, 2} = \frac{- ( - 11 ) \pm 1}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 11 ) + 1}{2 \cdot 3}, y_{2} = \frac{- ( - 11 ) - 1}{2 \cdot 3}

y = \frac{- ( - 11 ) + 1}{2 \cdot 3} : 2

y = \frac{- ( - 11 ) - 1}{2 \cdot 3} : \frac{5}{3}

The solutions to the quadratic equation are:

y = 2, y = \frac{5}{3}

13. 4^{2} = - 6. 7^{2} + (2 \cdot 3.75 x)

3 z \leq 9

5 x - 8 = 2 x + 4

25 - w^{2}

2 (3 a - 3) = 5 a - 2 a + 1

What are the solutions to the equation 3(1-y)^2+5(1-y)+2=0 ?
The solutions to the equation 3(1-y)^2+5(1-y)+2=0 are y=2,y= 5/3
Find the zeros of 3(1-y)^2+5(1-y)+2=0
The zeros of 3(1-y)^2+5(1-y)+2=0 are y=2,y= 5/3