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

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

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

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

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

x = 5, x = - 1

Solution steps

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

Expand

(x - 1)^{2} - 3 : x^{2} - 2 x - 2

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

Switch sides

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

Move

3

to the left side

x^{2} - 2 x - 5 = 2 x

Move

2 x

to the left side

x^{2} - 4 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

x_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 6}{2 \cdot 1}

x = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

x = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 5, x = - 1

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

3 x^{2} + 60 = 36 x

49 x^{2} + 70 x + 25 = 0

a^{2} + 3 a + 2 = 0

a x^{2} + b x + c = (5 x - 6) (2 x - 1)

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