What are the solutions to the equation 72=(8x+3)(8x-3) ?

The solutions to the equation 72=(8x+3)(8x-3) are x= 9/8 ,x=-9/8

Find the zeros of 72=(8x+3)(8x-3)

The zeros of 72=(8x+3)(8x-3) are x= 9/8 ,x=-9/8

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72 = (8 x + 3) (8 x - 3)

x = \frac{9}{8}, x = - \frac{9}{8}

Solution steps

72 = (8 x + 3) (8 x - 3)

Expand

(8 x + 3) (8 x - 3) : 64 x^{2} - 9

72 = 64 x^{2} - 9

Switch sides

64 x^{2} - 9 = 72

Move

72

to the left side

64 x^{2} - 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 64 ( - 81 )}{2 \cdot 64}

0^{2} - 4 \cdot 64 (- 81) = 144

x_{1, 2} = \frac{- 0 \pm 144}{2 \cdot 64}

Separate the solutions

x_{1} = \frac{- 0 + 144}{2 \cdot 64}, x_{2} = \frac{- 0 - 144}{2 \cdot 64}

x = \frac{- 0 + 144}{2 \cdot 64} : \frac{9}{8}

x = \frac{- 0 - 144}{2 \cdot 64} : - \frac{9}{8}

The solutions to the quadratic equation are:

x = \frac{9}{8}, x = - \frac{9}{8}

36 x^{2} - \frac{49}{64} = 0

x^{2} + 8 x + 96 = 0

y, x^{2} + y^{2} = c e^{2 x}

(6 x - 2)^{2} = 9

- x^{2} + 3 x - 28 = 0

What are the solutions to the equation 72=(8x+3)(8x-3) ?
The solutions to the equation 72=(8x+3)(8x-3) are x= 9/8 ,x=-9/8
Find the zeros of 72=(8x+3)(8x-3)
The zeros of 72=(8x+3)(8x-3) are x= 9/8 ,x=-9/8