What are the solutions to the equation (x-3)^2+1=2x-5 ?

The solutions to the equation (x-3)^2+1=2x-5 are x=5,x=3

Find the zeros of (x-3)^2+1=2x-5

The zeros of (x-3)^2+1=2x-5 are x=5,x=3

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(x - 3)^{2} + 1 = 2 x - 5

x = 5, x = 3

Solution steps

(x - 3)^{2} + 1 = 2 x - 5

Expand

(x - 3)^{2} + 1 : x^{2} - 6 x + 10

x^{2} - 6 x + 10 = 2 x - 5

Move

5

to the left side

x^{2} - 6 x + 15 = 2 x

Move

2 x

to the left side

x^{2} - 8 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 15}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 15 = 2

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

Separate the solutions

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

x = \frac{- ( - 8 ) + 2}{2 \cdot 1} : 5

x = \frac{- ( - 8 ) - 2}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

x = 5, x = 3

\frac{1}{4} (2 A) + 2 A x + B + A x^{2} + B x + C = x^{2} - 2 x

0.48 x^{2} - 1.88 x - 1.68 = 0

7 x^{2} - 3 x + j = 0

z^{2} - 10 z = - 5

3 t^{2} - 21 t + 30 = 0

What are the solutions to the equation (x-3)^2+1=2x-5 ?
The solutions to the equation (x-3)^2+1=2x-5 are x=5,x=3
Find the zeros of (x-3)^2+1=2x-5
The zeros of (x-3)^2+1=2x-5 are x=5,x=3