What are the solutions to the equation (2x+1)^2+(2x+3)^2=290 ?

The solutions to the equation (2x+1)^2+(2x+3)^2=290 are x=5,x=-7

Find the zeros of (2x+1)^2+(2x+3)^2=290

The zeros of (2x+1)^2+(2x+3)^2=290 are x=5,x=-7

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(2 x + 1)^{2} + (2 x + 3)^{2} = 290

x = 5, x = - 7

Solution steps

(2 x + 1)^{2} + (2 x + 3)^{2} = 290

Expand

(2 x + 1)^{2} + (2 x + 3)^{2} : 8 x^{2} + 16 x + 10

8 x^{2} + 16 x + 10 = 290

Move

290

to the left side

8 x^{2} + 16 x - 280 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 8 ( - 280 )}{2 \cdot 8}

1 6^{2} - 4 \cdot 8 (- 280) = 96

x_{1, 2} = \frac{- 16 \pm 96}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 16 + 96}{2 \cdot 8}, x_{2} = \frac{- 16 - 96}{2 \cdot 8}

x = \frac{- 16 + 96}{2 \cdot 8} : 5

x = \frac{- 16 - 96}{2 \cdot 8} : - 7

The solutions to the quadratic equation are:

x = 5, x = - 7

5 x^{2} - 3 x - 36 = 0

9 x^{2} + 36 x + 36 = 0

7 x^{2} - 8 x - 11 = 0

6 (2 x - 3)^{2} - 24 = 0

4 (x - 5)^{2} = 20

What are the solutions to the equation (2x+1)^2+(2x+3)^2=290 ?
The solutions to the equation (2x+1)^2+(2x+3)^2=290 are x=5,x=-7
Find the zeros of (2x+1)^2+(2x+3)^2=290
The zeros of (2x+1)^2+(2x+3)^2=290 are x=5,x=-7