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

The solutions to the equation (2-3x)^2=x+1 are x=(13+sqrt(61))/(18),x=(13-sqrt(61))/(18)

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

The zeros of (2-3x)^2=x+1 are x=(13+sqrt(61))/(18),x=(13-sqrt(61))/(18)

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

x = \frac{13 + 61}{18}, x = \frac{13 - 61}{18}

Decimal Notation

x = 1.15612 \dots, x = 0.28831 \dots

Solution steps

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

Expand

(2 - 3 x)^{2} : 4 - 12 x + 9 x^{2}

4 - 12 x + 9 x^{2} = x + 1

Move

1

to the left side

9 x^{2} - 12 x + 3 = x

Move

x

to the left side

9 x^{2} - 13 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 9 \cdot 3}{2 \cdot 9}

(- 13)^{2} - 4 \cdot 9 \cdot 3 = 61

x_{1, 2} = \frac{- ( - 13 ) \pm 61}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 61}{2 \cdot 9}, x_{2} = \frac{- ( - 13 ) - 61}{2 \cdot 9}

x = \frac{- ( - 13 ) + 61}{2 \cdot 9} : \frac{13 + 61}{18}

x = \frac{- ( - 13 ) - 61}{2 \cdot 9} : \frac{13 - 61}{18}

The solutions to the quadratic equation are:

x = \frac{13 + 61}{18}, x = \frac{13 - 61}{18}

(x + 6)^{2} = 121

(x + 6)^{2} = 144

1600 = 85 x - 150 - x^{2}

2 x^{2} = 32 x

3 x^{2} - 6 x + 10 = 0

What are the solutions to the equation (2-3x)^2=x+1 ?
The solutions to the equation (2-3x)^2=x+1 are x=(13+sqrt(61))/(18),x=(13-sqrt(61))/(18)
Find the zeros of (2-3x)^2=x+1
The zeros of (2-3x)^2=x+1 are x=(13+sqrt(61))/(18),x=(13-sqrt(61))/(18)