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

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

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

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

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

x = 1

Solution steps

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

Expand

3 (x - 1) : 3 x - 3

x^{2} + x - 2 = 3 x - 3

Move

3

to the left side

x^{2} + x + 1 = 3 x

Move

3 x

to the left side

x^{2} - 2 x + 1 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

x_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

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

\frac{- ( - 2 )}{2 \cdot 1} = 1

x = 1

The solution to the quadratic equation is:

x = 1

x^{2} - 12 x + 36

40 X \cdot 1 = 40

\frac{n}{4} \leq 0

5 x = x + 8

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