What are the solutions to the equation (2x-4)^2-10(2x-4)+21=0 ?

The solutions to the equation (2x-4)^2-10(2x-4)+21=0 are x= 11/2 ,x= 7/2

Find the zeros of (2x-4)^2-10(2x-4)+21=0

The zeros of (2x-4)^2-10(2x-4)+21=0 are x= 11/2 ,x= 7/2

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(2 x - 4)^{2} - 10 (2 x - 4) + 21 = 0

x = \frac{11}{2}, x = \frac{7}{2}

Solution steps

(2 x - 4)^{2} - 10 (2 x - 4) + 21 = 0

Expand

(2 x - 4)^{2} - 10 (2 x - 4) + 21 : 4 x^{2} - 36 x + 77

4 x^{2} - 36 x + 77 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 4 \cdot 77}{2 \cdot 4}

(- 36)^{2} - 4 \cdot 4 \cdot 77 = 8

x_{1, 2} = \frac{- ( - 36 ) \pm 8}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 36 ) + 8}{2 \cdot 4}, x_{2} = \frac{- ( - 36 ) - 8}{2 \cdot 4}

x = \frac{- ( - 36 ) + 8}{2 \cdot 4} : \frac{11}{2}

x = \frac{- ( - 36 ) - 8}{2 \cdot 4} : \frac{7}{2}

The solutions to the quadratic equation are:

x = \frac{11}{2}, x = \frac{7}{2}

2 x^{2} = 3 x

\frac{16}{3} + \frac{16}{15} + \frac{16}{35}

\frac{6}{\frac{3}{4}}

\frac{1}{4} \cdot 2

15 x^{2} - 7 x - 2

What are the solutions to the equation (2x-4)^2-10(2x-4)+21=0 ?
The solutions to the equation (2x-4)^2-10(2x-4)+21=0 are x= 11/2 ,x= 7/2
Find the zeros of (2x-4)^2-10(2x-4)+21=0
The zeros of (2x-4)^2-10(2x-4)+21=0 are x= 11/2 ,x= 7/2