What are the solutions to the equation (y-3)^2-5(y-3)=-4 ?

The solutions to the equation (y-3)^2-5(y-3)=-4 are y=7,y=4

Find the zeros of (y-3)^2-5(y-3)=-4

The zeros of (y-3)^2-5(y-3)=-4 are y=7,y=4

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(y - 3)^{2} - 5 (y - 3) = - 4

y = 7, y = 4

Solution steps

(y - 3)^{2} - 5 (y - 3) = - 4

Expand

(y - 3)^{2} - 5 (y - 3) : y^{2} - 11 y + 24

y^{2} - 11 y + 24 = - 4

Move

4

to the left side

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

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

1 5^{0}

- \frac{3}{2} (2)^{- 1}

f (g (x))

What are the solutions to the equation (y-3)^2-5(y-3)=-4 ?
The solutions to the equation (y-3)^2-5(y-3)=-4 are y=7,y=4
Find the zeros of (y-3)^2-5(y-3)=-4
The zeros of (y-3)^2-5(y-3)=-4 are y=7,y=4