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

The solutions to the equation 2y^2-8y-3=0 are y=(4+sqrt(22))/2 ,y=(4-sqrt(22))/2

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

The zeros of 2y^2-8y-3=0 are y=(4+sqrt(22))/2 ,y=(4-sqrt(22))/2

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

y = \frac{4 + 22}{2}, y = \frac{4 - 22}{2}

Decimal

y = 4.34520 \dots, y = - 0.34520 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 2 (- 3) = 222

y_{1, 2} = \frac{- ( - 8 ) \pm 2 22}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 2 22}{2 \cdot 2}, y_{2} = \frac{- ( - 8 ) - 2 22}{2 \cdot 2}

y = \frac{- ( - 8 ) + 2 22}{2 \cdot 2} : \frac{4 + 22}{2}

y = \frac{- ( - 8 ) - 2 22}{2 \cdot 2} : \frac{4 - 22}{2}

The solutions to the quadratic equation are:

y = \frac{4 + 22}{2}, y = \frac{4 - 22}{2}

- 4 n (n - 2) = 6 (n + 3) - 11 n^{2}

2 (x + 2)^{2} = 72

- x^{2} + 7 x - 8 = 0

- x^{2} + 2 x = x + 4

x^{2} - 3 = x - 2

What are the solutions to the equation 2y^2-8y-3=0 ?
The solutions to the equation 2y^2-8y-3=0 are y=(4+sqrt(22))/2 ,y=(4-sqrt(22))/2
Find the zeros of 2y^2-8y-3=0
The zeros of 2y^2-8y-3=0 are y=(4+sqrt(22))/2 ,y=(4-sqrt(22))/2