What are the solutions to the equation 4y^2+20y-11=0 ?

The solutions to the equation 4y^2+20y-11=0 are y= 1/2 ,y=-11/2

Find the zeros of 4y^2+20y-11=0

The zeros of 4y^2+20y-11=0 are y= 1/2 ,y=-11/2

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4 y^{2} + 20 y - 11 = 0

y = \frac{1}{2}, y = - \frac{11}{2}

Solution steps

4 y^{2} + 20 y - 11 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 ( - 11 )}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 (- 11) = 24

y_{1, 2} = \frac{- 20 \pm 24}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 20 + 24}{2 \cdot 4}, y_{2} = \frac{- 20 - 24}{2 \cdot 4}

y = \frac{- 20 + 24}{2 \cdot 4} : \frac{1}{2}

y = \frac{- 20 - 24}{2 \cdot 4} : - \frac{11}{2}

The solutions to the quadratic equation are:

y = \frac{1}{2}, y = - \frac{11}{2}

- 6 - 2 r^{2} = - 26

2 x^{2} = 4 x - 6

1 - x^{2} = 0

5 (x + 3)^{2} - 7 = 68

4 x (x - 1) = 3

What are the solutions to the equation 4y^2+20y-11=0 ?
The solutions to the equation 4y^2+20y-11=0 are y= 1/2 ,y=-11/2
Find the zeros of 4y^2+20y-11=0
The zeros of 4y^2+20y-11=0 are y= 1/2 ,y=-11/2