What are the solutions to the equation 2y^2-2y-10=1 ?

The solutions to the equation 2y^2-2y-10=1 are y=(1+sqrt(23))/2 ,y=(1-sqrt(23))/2

Find the zeros of 2y^2-2y-10=1

The zeros of 2y^2-2y-10=1 are y=(1+sqrt(23))/2 ,y=(1-sqrt(23))/2

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2 y^{2} - 2 y - 10 = 1

y = \frac{1 + 23}{2}, y = \frac{1 - 23}{2}

Decimal Notation

y = 2.89791 \dots, y = - 1.89791 \dots

Solution steps

2 y^{2} - 2 y - 10 = 1

Move

1

to the left side

2 y^{2} - 2 y - 11 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 11) = 223

y_{1, 2} = \frac{- ( - 2 ) \pm 2 23}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 2 23}{2 \cdot 2}, y_{2} = \frac{- ( - 2 ) - 2 23}{2 \cdot 2}

y = \frac{- ( - 2 ) + 2 23}{2 \cdot 2} : \frac{1 + 23}{2}

y = \frac{- ( - 2 ) - 2 23}{2 \cdot 2} : \frac{1 - 23}{2}

The solutions to the quadratic equation are:

y = \frac{1 + 23}{2}, y = \frac{1 - 23}{2}

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

x^{2} - 17 x + 30 = 0

x^{2} - 2 x - 3 = 0

t^{2} + 1 = 0

(5 x - 3)^{2} = 49

What are the solutions to the equation 2y^2-2y-10=1 ?
The solutions to the equation 2y^2-2y-10=1 are y=(1+sqrt(23))/2 ,y=(1-sqrt(23))/2
Find the zeros of 2y^2-2y-10=1
The zeros of 2y^2-2y-10=1 are y=(1+sqrt(23))/2 ,y=(1-sqrt(23))/2