What are the solutions to the equation 4y^2-7y-4=0 ?

The solutions to the equation 4y^2-7y-4=0 are y=(7+sqrt(113))/8 ,y=(7-sqrt(113))/8

Find the zeros of 4y^2-7y-4=0

The zeros of 4y^2-7y-4=0 are y=(7+sqrt(113))/8 ,y=(7-sqrt(113))/8

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4 y^{2} - 7 y - 4 = 0

y = \frac{7 + 113}{8}, y = \frac{7 - 113}{8}

Decimal Notation

y = 2.20376 \dots, y = - 0.45376 \dots

Solution steps

4 y^{2} - 7 y - 4 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

(- 7)^{2} - 4 \cdot 4 (- 4) = 113

y_{1, 2} = \frac{- ( - 7 ) \pm 113}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 7 ) + 113}{2 \cdot 4}, y_{2} = \frac{- ( - 7 ) - 113}{2 \cdot 4}

y = \frac{- ( - 7 ) + 113}{2 \cdot 4} : \frac{7 + 113}{8}

y = \frac{- ( - 7 ) - 113}{2 \cdot 4} : \frac{7 - 113}{8}

The solutions to the quadratic equation are:

y = \frac{7 + 113}{8}, y = \frac{7 - 113}{8}

117 = 0.0443213 (x)^{2}

0 = 30 - \frac{1}{2} \cdot 9.8 \cdot t^{2}

5 x^{2} - 8 x - 6 = 0

- t^{2} + 2 t - 49 = 0

x, 4 x^{2} - 4 x y + 7 y^{2} + 12 x + 6 y - 9 = 0

What are the solutions to the equation 4y^2-7y-4=0 ?
The solutions to the equation 4y^2-7y-4=0 are y=(7+sqrt(113))/8 ,y=(7-sqrt(113))/8
Find the zeros of 4y^2-7y-4=0
The zeros of 4y^2-7y-4=0 are y=(7+sqrt(113))/8 ,y=(7-sqrt(113))/8