What are the solutions to the equation (x-3)^2-5(x-3)+4=0 ?

The solutions to the equation (x-3)^2-5(x-3)+4=0 are x=7,x=4

Find the zeros of (x-3)^2-5(x-3)+4=0

The zeros of (x-3)^2-5(x-3)+4=0 are x=7,x=4

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(x - 3)^{2} - 5 (x - 3) + 4 = 0

x = 7, x = 4

Solution steps

(x - 3)^{2} - 5 (x - 3) + 4 = 0

Expand

(x - 3)^{2} - 5 (x - 3) + 4 : x^{2} - 11 x + 28

x^{2} - 11 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 28 = 3

x_{1, 2} = \frac{- ( - 11 ) \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 11 ) - 3}{2 \cdot 1}

x = \frac{- ( - 11 ) + 3}{2 \cdot 1} : 7

x = \frac{- ( - 11 ) - 3}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 7, x = 4

2 x^{2} - 16 x + 24 = 0

\frac{3}{2} m^{2} + 43 m + 3 = 0

\frac{2}{3} x^{2} - x = 0

3 a^{2} + 4 a - 15 = 0

x^{2} + 10 x = 11

What are the solutions to the equation (x-3)^2-5(x-3)+4=0 ?
The solutions to the equation (x-3)^2-5(x-3)+4=0 are x=7,x=4
Find the zeros of (x-3)^2-5(x-3)+4=0
The zeros of (x-3)^2-5(x-3)+4=0 are x=7,x=4