What are the solutions to the equation (2x-3)^2-(x+5)^2=0 ?

The solutions to the equation (2x-3)^2-(x+5)^2=0 are x=8,x=-2/3

Find the zeros of (2x-3)^2-(x+5)^2=0

The zeros of (2x-3)^2-(x+5)^2=0 are x=8,x=-2/3

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(2 x - 3)^{2} - (x + 5)^{2} = 0

x = 8, x = - \frac{2}{3}

Decimal Notation

x = 8, x = - 0.66666 \dots

Solution steps

(2 x - 3)^{2} - (x + 5)^{2} = 0

Expand

(2 x - 3)^{2} - (x + 5)^{2} : 3 x^{2} - 22 x - 16

3 x^{2} - 22 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 3 ( - 16 )}{2 \cdot 3}

(- 22)^{2} - 4 \cdot 3 (- 16) = 26

x_{1, 2} = \frac{- ( - 22 ) \pm 26}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 22 ) + 26}{2 \cdot 3}, x_{2} = \frac{- ( - 22 ) - 26}{2 \cdot 3}

x = \frac{- ( - 22 ) + 26}{2 \cdot 3} : 8

x = \frac{- ( - 22 ) - 26}{2 \cdot 3} : - \frac{2}{3}

The solutions to the quadratic equation are:

x = 8, x = - \frac{2}{3}

0.5 x^{2} - 3 x = - x + 2.5

e = x^{2} - 3 x + 5

(10 - x) (\frac{x}{4 - 3}) = 2

\frac{x}{2} = \frac{( x ^{2} )}{7}

x, 2 x \cdot y^{1} + y = \frac{( 2 x ^{2} )}{( y ^{3} )}

What are the solutions to the equation (2x-3)^2-(x+5)^2=0 ?
The solutions to the equation (2x-3)^2-(x+5)^2=0 are x=8,x=-2/3
Find the zeros of (2x-3)^2-(x+5)^2=0
The zeros of (2x-3)^2-(x+5)^2=0 are x=8,x=-2/3