What are the solutions to the equation solvefor y,y^2=11y-28 ?

The solutions to the equation solvefor y,y^2=11y-28 are y=7,y=4

Find the zeros of solvefor y,y^2=11y-28

The zeros of solvefor y,y^2=11y-28 are y=7,y=4

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y, y^{2} = 11 y - 28

y = 7, y = 4

Solution steps

y^{2} = 11 y - 28

Move

28

to the left side

y^{2} + 28 = 11 y

Move

11 y

to the left side

y^{2} + 28 - 11 y = 0

Write in the standard form

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

y^{2} - 11 y + 28 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 7, y = 4

1 0^{2} + x^{2} = 1 5^{2}

x, x^{2} = 2 y^{2}

11 e^{2} + 5 = 7 e

12 x = 2 x^{2}

2 x^{2} + 13 = 15

What are the solutions to the equation solvefor y,y^2=11y-28 ?
The solutions to the equation solvefor y,y^2=11y-28 are y=7,y=4
Find the zeros of solvefor y,y^2=11y-28
The zeros of solvefor y,y^2=11y-28 are y=7,y=4