What are the solutions to the equation -3y^2+8y-3=0 ?

The solutions to the equation -3y^2+8y-3=0 are y=(4-sqrt(7))/3 ,y=(4+sqrt(7))/3

Find the zeros of -3y^2+8y-3=0

The zeros of -3y^2+8y-3=0 are y=(4-sqrt(7))/3 ,y=(4+sqrt(7))/3

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- 3 y^{2} + 8 y - 3 = 0

y = \frac{4 - 7}{3}, y = \frac{4 + 7}{3}

Decimal

y = 0.45141 \dots, y = 2.21525 \dots

Solution steps

- 3 y^{2} + 8 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 3 ) ( - 3 )}{2 ( - 3 )}

8^{2} - 4 (- 3) (- 3) = 27

y_{1, 2} = \frac{- 8 \pm 2 7}{2 ( - 3 )}

Separate the solutions

y_{1} = \frac{- 8 + 2 7}{2 ( - 3 )}, y_{2} = \frac{- 8 - 2 7}{2 ( - 3 )}

y = \frac{- 8 + 2 7}{2 ( - 3 )} : \frac{4 - 7}{3}

y = \frac{- 8 - 2 7}{2 ( - 3 )} : \frac{4 + 7}{3}

The solutions to the quadratic equation are:

y = \frac{4 - 7}{3}, y = \frac{4 + 7}{3}

x, f (x + 3) = x^{2}

(3 z - 18)^{2} + 42 = - 3

r^{2} = 8

4 x^{2} - 10 = 39

x^{2} - x + 6 = 8

What are the solutions to the equation -3y^2+8y-3=0 ?
The solutions to the equation -3y^2+8y-3=0 are y=(4-sqrt(7))/3 ,y=(4+sqrt(7))/3
Find the zeros of -3y^2+8y-3=0
The zeros of -3y^2+8y-3=0 are y=(4-sqrt(7))/3 ,y=(4+sqrt(7))/3