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

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

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

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

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y^{2} - 28 = 3 y

y = 7, y = - 4

Solution steps

y^{2} - 28 = 3 y

Move

3 y

to the left side

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

Write in the standard form

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

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

\frac{y}{3} - \frac{3}{4} > \frac{1}{2}

- 8 \leq 4 x - 7 \leq 6

\frac{3}{4} \cdot \frac{4}{7}

3 6^{\frac{5}{2}}

x (- 2 x - 13) = - 1

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