What are the solutions to the equation (y-1)^2-13(y-1)+36=0 ?

The solutions to the equation (y-1)^2-13(y-1)+36=0 are y=10,y=5

Find the zeros of (y-1)^2-13(y-1)+36=0

The zeros of (y-1)^2-13(y-1)+36=0 are y=10,y=5

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(y - 1)^{2} - 13 (y - 1) + 36 = 0

y = 10, y = 5

Solution steps

(y - 1)^{2} - 13 (y - 1) + 36 = 0

Expand

(y - 1)^{2} - 13 (y - 1) + 36 : y^{2} - 15 y + 50

y^{2} - 15 y + 50 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 1 \cdot 50}{2 \cdot 1}

(- 15)^{2} - 4 \cdot 1 \cdot 50 = 5

y_{1, 2} = \frac{- ( - 15 ) \pm 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 15 ) + 5}{2 \cdot 1}, y_{2} = \frac{- ( - 15 ) - 5}{2 \cdot 1}

y = \frac{- ( - 15 ) + 5}{2 \cdot 1} : 10

y = \frac{- ( - 15 ) - 5}{2 \cdot 1} : 5

The solutions to the quadratic equation are:

y = 10, y = 5

2 (x - 1) \geq 8

\frac{4 + 3 i}{3 - i}

∣ 3 x - 4 ∣ > 8

(3 x - 4)^{2} = 64

- (\frac{2}{5})^{2}

What are the solutions to the equation (y-1)^2-13(y-1)+36=0 ?
The solutions to the equation (y-1)^2-13(y-1)+36=0 are y=10,y=5
Find the zeros of (y-1)^2-13(y-1)+36=0
The zeros of (y-1)^2-13(y-1)+36=0 are y=10,y=5