What are the solutions to the equation -2y^2-5y+6=0 ?

The solutions to the equation -2y^2-5y+6=0 are y=-(5+sqrt(73))/4 ,y=(sqrt(73)-5)/4

Find the zeros of -2y^2-5y+6=0

The zeros of -2y^2-5y+6=0 are y=-(5+sqrt(73))/4 ,y=(sqrt(73)-5)/4

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- 2 y^{2} - 5 y + 6 = 0

y = - \frac{5 + 73}{4}, y = \frac{73 - 5}{4}

Decimal Notation

y = - 3.38600 \dots, y = 0.88600 \dots

Solution steps

- 2 y^{2} - 5 y + 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) \cdot 6}{2 ( - 2 )}

(- 5)^{2} - 4 (- 2) \cdot 6 = 73

y_{1, 2} = \frac{- ( - 5 ) \pm 73}{2 ( - 2 )}

Separate the solutions

y_{1} = \frac{- ( - 5 ) + 73}{2 ( - 2 )}, y_{2} = \frac{- ( - 5 ) - 73}{2 ( - 2 )}

y = \frac{- ( - 5 ) + 73}{2 ( - 2 )} : - \frac{5 + 73}{4}

y = \frac{- ( - 5 ) - 73}{2 ( - 2 )} : \frac{73 - 5}{4}

The solutions to the quadratic equation are:

y = - \frac{5 + 73}{4}, y = \frac{73 - 5}{4}

3 s^{2} + 0.5 s + 2 = 0

- x^{2} + 0 x + 1 = 0

3 x (2 x - 8) = 0

x^{2} - 19 x - 722 = 0

x^{2} + y^{2} = 1, x

What are the solutions to the equation -2y^2-5y+6=0 ?
The solutions to the equation -2y^2-5y+6=0 are y=-(5+sqrt(73))/4 ,y=(sqrt(73)-5)/4
Find the zeros of -2y^2-5y+6=0
The zeros of -2y^2-5y+6=0 are y=-(5+sqrt(73))/4 ,y=(sqrt(73)-5)/4