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

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

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

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

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

y = 1, y = - 4

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} : 1

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

The solutions to the quadratic equation are:

y = 1, y = - 4

\frac{1200 - n}{1200} = 0.04

- 8 = \frac{x + 11}{- 2}

- 11 u - 8 + 3 u = 24

∣ 3 x - 2 ∣ \geq 5

22 x - 6 + 70 = 148

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