What are the solutions to the equation (2x-5)(2x+5)-119=0 ?

The solutions to the equation (2x-5)(2x+5)-119=0 are x=6,x=-6

Find the zeros of (2x-5)(2x+5)-119=0

The zeros of (2x-5)(2x+5)-119=0 are x=6,x=-6

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(2 x - 5) (2 x + 5) - 119 = 0

x = 6, x = - 6

Solution steps

(2 x - 5) (2 x + 5) - 119 = 0

Expand

(2 x - 5) (2 x + 5) - 119 : 4 x^{2} - 144

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

x^{2} - 10 x + 28 = 0

y, y^{2} = x^{2}

r^{2} = 9

16 x^{2} + 51 = 56 x

4 q (q + 2) + 10 = 0

What are the solutions to the equation (2x-5)(2x+5)-119=0 ?
The solutions to the equation (2x-5)(2x+5)-119=0 are x=6,x=-6
Find the zeros of (2x-5)(2x+5)-119=0
The zeros of (2x-5)(2x+5)-119=0 are x=6,x=-6