What are the solutions to the equation (3x-5)^2-(2x+1)^2=24 ?

The solutions to the equation (3x-5)^2-(2x+1)^2=24 are x= 34/5 ,x=0

Find the zeros of (3x-5)^2-(2x+1)^2=24

The zeros of (3x-5)^2-(2x+1)^2=24 are x= 34/5 ,x=0

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(3 x - 5)^{2} - (2 x + 1)^{2} = 24

x = \frac{34}{5}, x = 0

Decimal

x = 6.8, x = 0

Solution steps

(3 x - 5)^{2} - (2 x + 1)^{2} = 24

Expand

(3 x - 5)^{2} - (2 x + 1)^{2} : 5 x^{2} - 34 x + 24

5 x^{2} - 34 x + 24 = 24

Move

24

to the left side

5 x^{2} - 34 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 34 ) \pm ( - 34 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

(- 34)^{2} - 4 \cdot 5 \cdot 0 = 34

x_{1, 2} = \frac{- ( - 34 ) \pm 34}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 34 ) + 34}{2 \cdot 5}, x_{2} = \frac{- ( - 34 ) - 34}{2 \cdot 5}

x = \frac{- ( - 34 ) + 34}{2 \cdot 5} : \frac{34}{5}

x = \frac{- ( - 34 ) - 34}{2 \cdot 5} : 0

The solutions to the quadratic equation are:

x = \frac{34}{5}, x = 0

x, f = a (x - h)^{2} + k

4 n^{2} - 9 = - n

x, (x - h)^{2} + (y - k)^{2} = r^{2}

4 w^{2} + 12 w - 44 = 0

0 = x^{2} + 5 x + 8

What are the solutions to the equation (3x-5)^2-(2x+1)^2=24 ?
The solutions to the equation (3x-5)^2-(2x+1)^2=24 are x= 34/5 ,x=0
Find the zeros of (3x-5)^2-(2x+1)^2=24
The zeros of (3x-5)^2-(2x+1)^2=24 are x= 34/5 ,x=0