What are the solutions to the equation (3x+1)^2+(x+2)^2=65 ?

The solutions to the equation (3x+1)^2+(x+2)^2=65 are x=2,x=-3

Find the zeros of (3x+1)^2+(x+2)^2=65

The zeros of (3x+1)^2+(x+2)^2=65 are x=2,x=-3

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(3 x + 1)^{2} + (x + 2)^{2} = 65

x = 2, x = - 3

Solution steps

(3 x + 1)^{2} + (x + 2)^{2} = 65

Expand

(3 x + 1)^{2} + (x + 2)^{2} : 10 x^{2} + 10 x + 5

10 x^{2} + 10 x + 5 = 65

Move

65

to the left side

10 x^{2} + 10 x - 60 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 10 ( - 60 )}{2 \cdot 10}

1 0^{2} - 4 \cdot 10 (- 60) = 50

x_{1, 2} = \frac{- 10 \pm 50}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- 10 + 50}{2 \cdot 10}, x_{2} = \frac{- 10 - 50}{2 \cdot 10}

x = \frac{- 10 + 50}{2 \cdot 10} : 2

x = \frac{- 10 - 50}{2 \cdot 10} : - 3

The solutions to the quadratic equation are:

x = 2, x = - 3

2 a^{2} - 5 a + 3 = 0

x^{2} - 7 x + 6 = 0

6 x^{2} = 121

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

24 x^{2} - 6 = 0

What are the solutions to the equation (3x+1)^2+(x+2)^2=65 ?
The solutions to the equation (3x+1)^2+(x+2)^2=65 are x=2,x=-3
Find the zeros of (3x+1)^2+(x+2)^2=65
The zeros of (3x+1)^2+(x+2)^2=65 are x=2,x=-3