What are the solutions to the equation (y+2)^2-14(y+2)=-24 ?

The solutions to the equation (y+2)^2-14(y+2)=-24 are y=10,y=0

Find the zeros of (y+2)^2-14(y+2)=-24

The zeros of (y+2)^2-14(y+2)=-24 are y=10,y=0

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(y + 2)^{2} - 14 (y + 2) = - 24

y = 10, y = 0

Solution steps

(y + 2)^{2} - 14 (y + 2) = - 24

Expand

(y + 2)^{2} - 14 (y + 2) : y^{2} - 10 y - 24

y^{2} - 10 y - 24 = - 24

Move

24

to the left side

y^{2} - 10 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 0 = 10

y_{1, 2} = \frac{- ( - 10 ) \pm 10}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 10 ) + 10}{2 \cdot 1}, y_{2} = \frac{- ( - 10 ) - 10}{2 \cdot 1}

y = \frac{- ( - 10 ) + 10}{2 \cdot 1} : 10

y = \frac{- ( - 10 ) - 10}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

y = 10, y = 0

3 b - 3

x^{2} + 3 x = - 1

- 5 (2 y - 15) + 4 y = 3

4 x^{2} + 28 x + 48

2 x^{4} + 3 x^{3} - 9 x^{2} + x + 3

What are the solutions to the equation (y+2)^2-14(y+2)=-24 ?
The solutions to the equation (y+2)^2-14(y+2)=-24 are y=10,y=0
Find the zeros of (y+2)^2-14(y+2)=-24
The zeros of (y+2)^2-14(y+2)=-24 are y=10,y=0