What are the solutions to the equation (1/2 x+6)^2-(x-3)^2=0 ?

The solutions to the equation (1/2 x+6)^2-(x-3)^2=0 are x=-2,x=18

Find the zeros of (1/2 x+6)^2-(x-3)^2=0

The zeros of (1/2 x+6)^2-(x-3)^2=0 are x=-2,x=18

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(\frac{1}{2} x + 6)^{2} - (x - 3)^{2} = 0

x = - 2, x = 18

Solution steps

(\frac{1}{2} x + 6)^{2} - (x - 3)^{2} = 0

Expand

(\frac{1}{2} x + 6)^{2} - (x - 3)^{2} : - \frac{3}{4} x^{2} + 12 x + 27

- \frac{3}{4} x^{2} + 12 x + 27 = 0

Write in the standard form

a x^{2} + b x + c = 0

- \frac{3 x ^{2}}{4} + 12 x + 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - \frac{3}{4} ) \cdot 27}{2 ( - \frac{3}{4} )}

1 2^{2} - 4 (- \frac{3}{4}) \cdot 27 = 15

x_{1, 2} = \frac{- 12 \pm 15}{2 ( - \frac{3}{4} )}

Separate the solutions

x_{1} = \frac{- 12 + 15}{2 ( - \frac{3}{4} )}, x_{2} = \frac{- 12 - 15}{2 ( - \frac{3}{4} )}

x = \frac{- 12 + 15}{2 ( - \frac{3}{4} )} : - 2

x = \frac{- 12 - 15}{2 ( - \frac{3}{4} )} : 18

The solutions to the quadratic equation are:

x = - 2, x = 18

3 x^{2} - 3 x + 4 = 0

(- 1 - 2 i) - (3 + 7 i)

f + 17 < 3

- 9 n = 10 - 8 n

6 x^{2} = - 12 x - 4

What are the solutions to the equation (1/2 x+6)^2-(x-3)^2=0 ?
The solutions to the equation (1/2 x+6)^2-(x-3)^2=0 are x=-2,x=18
Find the zeros of (1/2 x+6)^2-(x-3)^2=0
The zeros of (1/2 x+6)^2-(x-3)^2=0 are x=-2,x=18