What are the solutions to the equation 2y^2-4y-5=0 ?

The solutions to the equation 2y^2-4y-5=0 are y=(2+sqrt(14))/2 ,y=(2-sqrt(14))/2

Find the zeros of 2y^2-4y-5=0

The zeros of 2y^2-4y-5=0 are y=(2+sqrt(14))/2 ,y=(2-sqrt(14))/2

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2 y^{2} - 4 y - 5 = 0

y = \frac{2 + 14}{2}, y = \frac{2 - 14}{2}

Decimal Notation

y = 2.87082 \dots, y = - 0.87082 \dots

Solution steps

2 y^{2} - 4 y - 5 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 (- 5) = 214

y_{1, 2} = \frac{- ( - 4 ) \pm 2 14}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 2 14}{2 \cdot 2}, y_{2} = \frac{- ( - 4 ) - 2 14}{2 \cdot 2}

y = \frac{- ( - 4 ) + 2 14}{2 \cdot 2} : \frac{2 + 14}{2}

y = \frac{- ( - 4 ) - 2 14}{2 \cdot 2} : \frac{2 - 14}{2}

The solutions to the quadratic equation are:

y = \frac{2 + 14}{2}, y = \frac{2 - 14}{2}

\frac{9 ^{7}}{9 ^{2}}

4 (0.5 f - 0.25) = 6 + f

lo g_{2} (8) = y

1.0 6^{4}

(t^{\frac{1}{2}})^{2}

What are the solutions to the equation 2y^2-4y-5=0 ?
The solutions to the equation 2y^2-4y-5=0 are y=(2+sqrt(14))/2 ,y=(2-sqrt(14))/2
Find the zeros of 2y^2-4y-5=0
The zeros of 2y^2-4y-5=0 are y=(2+sqrt(14))/2 ,y=(2-sqrt(14))/2