What are the solutions to the equation (1+3x)^2=5x+9 ?

The solutions to the equation (1+3x)^2=5x+9 are x= 8/9 ,x=-1

Find the zeros of (1+3x)^2=5x+9

The zeros of (1+3x)^2=5x+9 are x= 8/9 ,x=-1

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(1 + 3 x)^{2} = 5 x + 9

x = \frac{8}{9}, x = - 1

Decimal Notation

x = 0.88888 \dots, x = - 1

Solution steps

(1 + 3 x)^{2} = 5 x + 9

Expand

(1 + 3 x)^{2} : 1 + 6 x + 9 x^{2}

1 + 6 x + 9 x^{2} = 5 x + 9

Move

9

to the left side

9 x^{2} + 6 x - 8 = 5 x

Move

5 x

to the left side

9 x^{2} + x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 9 ( - 8 )}{2 \cdot 9}

1^{2} - 4 \cdot 9 (- 8) = 17

x_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 1 + 17}{2 \cdot 9}, x_{2} = \frac{- 1 - 17}{2 \cdot 9}

x = \frac{- 1 + 17}{2 \cdot 9} : \frac{8}{9}

x = \frac{- 1 - 17}{2 \cdot 9} : - 1

The solutions to the quadratic equation are:

x = \frac{8}{9}, x = - 1

5 x^{2} - 25 x - 20 = 0

x (x + 3) - 3 = 0

a^{2} + 7 a - 8 = 0

b, y^{2} (2 x + b) = b^{2}

3 x^{2} + 6 x - 78 = 0

What are the solutions to the equation (1+3x)^2=5x+9 ?
The solutions to the equation (1+3x)^2=5x+9 are x= 8/9 ,x=-1
Find the zeros of (1+3x)^2=5x+9
The zeros of (1+3x)^2=5x+9 are x= 8/9 ,x=-1