What are the solutions to the equation 4y^2+11y-3=0 ?

The solutions to the equation 4y^2+11y-3=0 are y= 1/4 ,y=-3

Find the zeros of 4y^2+11y-3=0

The zeros of 4y^2+11y-3=0 are y= 1/4 ,y=-3

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

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

Decimal Notation

y = 0.25, y = - 3

Solution steps

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

Solve with the quadratic formula

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

1 1^{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} : \frac{1}{4}

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

The solutions to the quadratic equation are:

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

4 (x + 3) = 3 x + 2 + x

\frac{1}{e ^{- 1}}

\frac{8 a ^{3} - b ^{3}}{4 a ^{2} - b ^{2}}

\frac{3}{4} x^{2} - 5 = 16

x + 1 \leq 6 and \frac{1}{3} x < - 2

What are the solutions to the equation 4y^2+11y-3=0 ?
The solutions to the equation 4y^2+11y-3=0 are y= 1/4 ,y=-3
Find the zeros of 4y^2+11y-3=0
The zeros of 4y^2+11y-3=0 are y= 1/4 ,y=-3