What are the solutions to the equation y^2-3y-28=0 ?

The solutions to the equation y^2-3y-28=0 are y=7,y=-4

Find the zeros of y^2-3y-28=0

The zeros of y^2-3y-28=0 are y=7,y=-4

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y^{2} - 3 y - 28 = 0

y = 7, y = - 4

Solution steps

y^{2} - 3 y - 28 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 28) = 11

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

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 11}{2 \cdot 1}, y_{2} = \frac{- ( - 3 ) - 11}{2 \cdot 1}

y = \frac{- ( - 3 ) + 11}{2 \cdot 1} : 7

y = \frac{- ( - 3 ) - 11}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

y = 7, y = - 4

x, (x - 1) (3 x + 2 y - 1) = C

x^{2} - 4 x + 5 = 0

x (x + 8) = 3

6 x^{2} + x (\frac{5 - 2 x}{3}) + 3 x - 9 = 0

- 5 x^{2} + 250 = 0

What are the solutions to the equation y^2-3y-28=0 ?
The solutions to the equation y^2-3y-28=0 are y=7,y=-4
Find the zeros of y^2-3y-28=0
The zeros of y^2-3y-28=0 are y=7,y=-4