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

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

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

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

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

x = - 5, x = - \frac{1}{3}

Decimal Notation

x = - 5, x = - 0.33333 \dots

Solution steps

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

Expand

(x - 2)^{2} - (2 x + 3)^{2} : - 3 x^{2} - 16 x - 5

- 3 x^{2} - 16 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 ( - 3 ) ( - 5 )}{2 ( - 3 )}

(- 16)^{2} - 4 (- 3) (- 5) = 14

x_{1, 2} = \frac{- ( - 16 ) \pm 14}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 14}{2 ( - 3 )}, x_{2} = \frac{- ( - 16 ) - 14}{2 ( - 3 )}

x = \frac{- ( - 16 ) + 14}{2 ( - 3 )} : - 5

x = \frac{- ( - 16 ) - 14}{2 ( - 3 )} : - \frac{1}{3}

The solutions to the quadratic equation are:

x = - 5, x = - \frac{1}{3}

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

n^{2} + 9 n + 14 = 0

x^{2} + 3.9 x + 6.8 = 0

\frac{a + 3 x ^{2}}{b} = x^{2}

2 x^{2} + 2 x - 1

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