What are the solutions to the equation y^2+12=7y ?

The solutions to the equation y^2+12=7y are y=4,y=3

Find the zeros of y^2+12=7y

The zeros of y^2+12=7y are y=4,y=3

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y^{2} + 12 = 7 y

y = 4, y = 3

Solution steps

y^{2} + 12 = 7 y

Move

7 y

to the left side

y^{2} + 12 - 7 y = 0

Write in the standard form

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

y^{2} - 7 y + 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 12 = 1

y_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 1}, y_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 1}

y = \frac{- ( - 7 ) + 1}{2 \cdot 1} : 4

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

The solutions to the quadratic equation are:

y = 4, y = 3

4 x^{2} - 28 x + 49 = 5

w^{2} + 3 w = 3

4 x^{2} + 4 = - 7 x

2 X^{2} + 4 X - 6 = 0

8 x^{2} = - 4 x

What are the solutions to the equation y^2+12=7y ?
The solutions to the equation y^2+12=7y are y=4,y=3
Find the zeros of y^2+12=7y
The zeros of y^2+12=7y are y=4,y=3