What are the solutions to the equation 135=(2x+3)(2x-3) ?

The solutions to the equation 135=(2x+3)(2x-3) are x=6,x=-6

Find the zeros of 135=(2x+3)(2x-3)

The zeros of 135=(2x+3)(2x-3) are x=6,x=-6

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135 = (2 x + 3) (2 x - 3)

x = 6, x = - 6

Solution steps

135 = (2 x + 3) (2 x - 3)

Expand

(2 x + 3) (2 x - 3) : 4 x^{2} - 9

135 = 4 x^{2} - 9

Switch sides

4 x^{2} - 9 = 135

Move

135

to the left side

4 x^{2} - 144 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 4 ( - 144 )}{2 \cdot 4}

0^{2} - 4 \cdot 4 (- 144) = 48

x_{1, 2} = \frac{- 0 \pm 48}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 0 + 48}{2 \cdot 4}, x_{2} = \frac{- 0 - 48}{2 \cdot 4}

x = \frac{- 0 + 48}{2 \cdot 4} : 6

x = \frac{- 0 - 48}{2 \cdot 4} : - 6

The solutions to the quadratic equation are:

x = 6, x = - 6

9 x^{2} - 3 x = 0

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

y^{2} + y - 20 = 0

5 b^{2} + 3 b - 4 = 0

x^{2} - 13 x + 110 = 160 - 9 x

What are the solutions to the equation 135=(2x+3)(2x-3) ?
The solutions to the equation 135=(2x+3)(2x-3) are x=6,x=-6
Find the zeros of 135=(2x+3)(2x-3)
The zeros of 135=(2x+3)(2x-3) are x=6,x=-6