What are the solutions to the equation 4y^2-3=11y ?

The solutions to the equation 4y^2-3=11y are y=3,y=-1/4

Find the zeros of 4y^2-3=11y

The zeros of 4y^2-3=11y are y=3,y=-1/4

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4 y^{2} - 3 = 11 y

y = 3, y = - \frac{1}{4}

Decimal Notation

y = 3, y = - 0.25

Solution steps

4 y^{2} - 3 = 11 y

Move

11 y

to the left side

4 y^{2} - 3 - 11 y = 0

Write in the standard form

a x^{2} + b x + c = 0

4 y^{2} - 11 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

(- 11)^{2} - 4 \cdot 4 (- 3) = 13

y_{1, 2} = \frac{- ( - 11 ) \pm 13}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 11 ) + 13}{2 \cdot 4}, y_{2} = \frac{- ( - 11 ) - 13}{2 \cdot 4}

y = \frac{- ( - 11 ) + 13}{2 \cdot 4} : 3

y = \frac{- ( - 11 ) - 13}{2 \cdot 4} : - \frac{1}{4}

The solutions to the quadratic equation are:

y = 3, y = - \frac{1}{4}

2 v^{2} - 9 v = 5

0 = 64 + 48 t - 16 t^{2}

y, x = 3 y^{2} + k y

x, x^{2} + y^{2} = \frac{( z - 4 ) ^{2}}{4}

s^{2} - 22 s = 31

What are the solutions to the equation 4y^2-3=11y ?
The solutions to the equation 4y^2-3=11y are y=3,y=-1/4
Find the zeros of 4y^2-3=11y
The zeros of 4y^2-3=11y are y=3,y=-1/4