What are the solutions to the equation (3x-1)^2-(2x+3)^2=0 ?

The solutions to the equation (3x-1)^2-(2x+3)^2=0 are x=4,x=-2/5

Find the zeros of (3x-1)^2-(2x+3)^2=0

The zeros of (3x-1)^2-(2x+3)^2=0 are x=4,x=-2/5

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(3 x - 1)^{2} - (2 x + 3)^{2} = 0

x = 4, x = - \frac{2}{5}

Decimal Notation

x = 4, x = - 0.4

Solution steps

(3 x - 1)^{2} - (2 x + 3)^{2} = 0

Expand

(3 x - 1)^{2} - (2 x + 3)^{2} : 5 x^{2} - 18 x - 8

5 x^{2} - 18 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 5 ( - 8 )}{2 \cdot 5}

(- 18)^{2} - 4 \cdot 5 (- 8) = 22

x_{1, 2} = \frac{- ( - 18 ) \pm 22}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 22}{2 \cdot 5}, x_{2} = \frac{- ( - 18 ) - 22}{2 \cdot 5}

x = \frac{- ( - 18 ) + 22}{2 \cdot 5} : 4

x = \frac{- ( - 18 ) - 22}{2 \cdot 5} : - \frac{2}{5}

The solutions to the quadratic equation are:

x = 4, x = - \frac{2}{5}

(x - a)^{2} = b

y, y (y + 1) = x (x + 1) (x + 2)

3 m^{2} + 2 = 4, m^{2} + 23 m^{2} - 4 m^{2} = 2 - 2

2 x^{2} - 5 x + 2 = 0, x - 2 \geq 0

a = 4 z^{2} + 23 z - 35

What are the solutions to the equation (3x-1)^2-(2x+3)^2=0 ?
The solutions to the equation (3x-1)^2-(2x+3)^2=0 are x=4,x=-2/5
Find the zeros of (3x-1)^2-(2x+3)^2=0
The zeros of (3x-1)^2-(2x+3)^2=0 are x=4,x=-2/5