What are the solutions to the equation (y-5)^2-11(y-5)+24=0 ?

The solutions to the equation (y-5)^2-11(y-5)+24=0 are y=13,y=8

Find the zeros of (y-5)^2-11(y-5)+24=0

The zeros of (y-5)^2-11(y-5)+24=0 are y=13,y=8

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(y - 5)^{2} - 11 (y - 5) + 24 = 0

y = 13, y = 8

Solution steps

(y - 5)^{2} - 11 (y - 5) + 24 = 0

Expand

(y - 5)^{2} - 11 (y - 5) + 24 : y^{2} - 21 y + 104

y^{2} - 21 y + 104 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 1 \cdot 104}{2 \cdot 1}

(- 21)^{2} - 4 \cdot 1 \cdot 104 = 5

y_{1, 2} = \frac{- ( - 21 ) \pm 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 21 ) + 5}{2 \cdot 1}, y_{2} = \frac{- ( - 21 ) - 5}{2 \cdot 1}

y = \frac{- ( - 21 ) + 5}{2 \cdot 1} : 13

y = \frac{- ( - 21 ) - 5}{2 \cdot 1} : 8

The solutions to the quadratic equation are:

y = 13, y = 8

2 x^{2} + 7 x + 15 = x

105 x^{2} + 80 = 200 x

(9 z + 5)^{2} = 0

5 a^{2} - 6 a + 10 = 0

2 x^{2} - 26 x + 72 = 0

What are the solutions to the equation (y-5)^2-11(y-5)+24=0 ?
The solutions to the equation (y-5)^2-11(y-5)+24=0 are y=13,y=8
Find the zeros of (y-5)^2-11(y-5)+24=0
The zeros of (y-5)^2-11(y-5)+24=0 are y=13,y=8