What are the solutions to the equation (x+1)^2=3x+7 ?

The solutions to the equation (x+1)^2=3x+7 are x=3,x=-2

Find the zeros of (x+1)^2=3x+7

The zeros of (x+1)^2=3x+7 are x=3,x=-2

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(x + 1)^{2} = 3 x + 7

x = 3, x = - 2

Solution steps

(x + 1)^{2} = 3 x + 7

Expand

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

x^{2} + 2 x + 1 = 3 x + 7

Move

7

to the left side

x^{2} + 2 x - 6 = 3 x

Move

3 x

to the left side

x^{2} - x - 6 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

x = \frac{- ( - 1 ) + 5}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

x = 3, x = - 2

(x - 2) (x + 4) = 7

(x + 2)^{2} - 64 = 0

2 p = \frac{- 5 p - 6}{3}

3^{3}

6.5 (x - 6) = 19.5

What are the solutions to the equation (x+1)^2=3x+7 ?
The solutions to the equation (x+1)^2=3x+7 are x=3,x=-2
Find the zeros of (x+1)^2=3x+7
The zeros of (x+1)^2=3x+7 are x=3,x=-2