What are the solutions to the equation solvefor y,y^2-3y-4=0 ?

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

Find the zeros of solvefor y,y^2-3y-4=0

The zeros of solvefor y,y^2-3y-4=0 are y=4,y=-1

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

y = 4, y = - 1

Solution steps

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

y_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, y_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

y = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

y = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

y = 4, y = - 1

2 m^{2} - 7 m = 15

11 - 4 x^{2} = 187

x^{2} - x - 42 = 0

x^{2} + 3 = 1

x^{2} - 4 x - 20 = - 5 x

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