What are the solutions to the equation (2x-5)^2=(x-6)^2+(2x+1)(2x-9) ?

The solutions to the equation (2x-5)^2=(x-6)^2+(2x+1)(2x-9) are x=4-sqrt(14),x=4+sqrt(14)

Find the zeros of (2x-5)^2=(x-6)^2+(2x+1)(2x-9)

The zeros of (2x-5)^2=(x-6)^2+(2x+1)(2x-9) are x=4-sqrt(14),x=4+sqrt(14)

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(2 x - 5)^{2} = (x - 6)^{2} + (2 x + 1) (2 x - 9)

x = 4 - 14, x = 4 + 14

Decimal Notation

x = 0.25834 \dots, x = 7.74165 \dots

Solution steps

(2 x - 5)^{2} = (x - 6)^{2} + (2 x + 1) (2 x - 9)

Expand

(2 x - 5)^{2} : 4 x^{2} - 20 x + 25

Expand

(x - 6)^{2} + (2 x + 1) (2 x - 9) : 5 x^{2} - 28 x + 27

4 x^{2} - 20 x + 25 = 5 x^{2} - 28 x + 27

Move

27

to the left side

4 x^{2} - 20 x - 2 = 5 x^{2} - 28 x

Move

28 x

to the left side

4 x^{2} + 8 x - 2 = 5 x^{2}

Move

5 x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

8^{2} - 4 (- 1) (- 2) = 214

x_{1, 2} = \frac{- 8 \pm 2 14}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 8 + 2 14}{2 ( - 1 )}, x_{2} = \frac{- 8 - 2 14}{2 ( - 1 )}

x = \frac{- 8 + 2 14}{2 ( - 1 )} : 4 - 14

x = \frac{- 8 - 2 14}{2 ( - 1 )} : 4 + 14

The solutions to the quadratic equation are:

x = 4 - 14, x = 4 + 14

x^{2} = \frac{1}{100}

2 x^{2} - 7 x + 3 = 0

a, (a + 2 b) (a - b) = 10

4 x^{2} + 10 x + 3 = 0

(3 x + 6)^{2} = 100

What are the solutions to the equation (2x-5)^2=(x-6)^2+(2x+1)(2x-9) ?
The solutions to the equation (2x-5)^2=(x-6)^2+(2x+1)(2x-9) are x=4-sqrt(14),x=4+sqrt(14)
Find the zeros of (2x-5)^2=(x-6)^2+(2x+1)(2x-9)
The zeros of (2x-5)^2=(x-6)^2+(2x+1)(2x-9) are x=4-sqrt(14),x=4+sqrt(14)