What are the solutions to the equation (x-1)^2=(2x+6) ?

The solutions to the equation (x-1)^2=(2x+6) are x=5,x=-1

Find the zeros of (x-1)^2=(2x+6)

The zeros of (x-1)^2=(2x+6) are x=5,x=-1

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(x - 1)^{2} = (2 x + 6)

x = 5, x = - 1

Solution steps

(x - 1)^{2} = (2 x + 6)

Expand

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

x^{2} - 2 x + 1 = 2 x + 6

Move

6

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

∣ 9 - 2 x ∣ \geq ∣ 4 x ∣

2 t + 2 > 10

∣ 2 x + 8 ∣ > 6

\frac{x}{3} = \frac{2}{5}

∣ 7 x - 1 ∣ > 3

What are the solutions to the equation (x-1)^2=(2x+6) ?
The solutions to the equation (x-1)^2=(2x+6) are x=5,x=-1
Find the zeros of (x-1)^2=(2x+6)
The zeros of (x-1)^2=(2x+6) are x=5,x=-1