What are the solutions to the equation (2x)^2+(2x+3)^2=369 ?

The solutions to the equation (2x)^2+(2x+3)^2=369 are x=6,x=-15/2

Find the zeros of (2x)^2+(2x+3)^2=369

The zeros of (2x)^2+(2x+3)^2=369 are x=6,x=-15/2

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(2 x)^{2} + (2 x + 3)^{2} = 369

x = 6, x = - \frac{15}{2}

Decimal Notation

x = 6, x = - 7.5

Solution steps

(2 x)^{2} + (2 x + 3)^{2} = 369

Expand

(2 x)^{2} + (2 x + 3)^{2} : 8 x^{2} + 12 x + 9

8 x^{2} + 12 x + 9 = 369

Move

369

to the left side

8 x^{2} + 12 x - 360 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 8 ( - 360 )}{2 \cdot 8}

1 2^{2} - 4 \cdot 8 (- 360) = 108

x_{1, 2} = \frac{- 12 \pm 108}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 12 + 108}{2 \cdot 8}, x_{2} = \frac{- 12 - 108}{2 \cdot 8}

x = \frac{- 12 + 108}{2 \cdot 8} : 6

x = \frac{- 12 - 108}{2 \cdot 8} : - \frac{15}{2}

The solutions to the quadratic equation are:

x = 6, x = - \frac{15}{2}

4 (x - 2)^{2} + 1 = 21

x^{2} - 2 (m - 2) x + (m^{2} - 2 m - 3) = 0

x^{2} + 16 x - 225 = 0

x, (p - 1) x^{2} + 2 p x + p - 1 = 0

m = x^{2} a + x^{2} b

What are the solutions to the equation (2x)^2+(2x+3)^2=369 ?
The solutions to the equation (2x)^2+(2x+3)^2=369 are x=6,x=-15/2
Find the zeros of (2x)^2+(2x+3)^2=369
The zeros of (2x)^2+(2x+3)^2=369 are x=6,x=-15/2