What are the solutions to the equation (x+1)^2-2(x-2)^2=8 ?

The solutions to the equation (x+1)^2-2(x-2)^2=8 are x=5-sqrt(10),x=5+sqrt(10)

Find the zeros of (x+1)^2-2(x-2)^2=8

The zeros of (x+1)^2-2(x-2)^2=8 are x=5-sqrt(10),x=5+sqrt(10)

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(x + 1)^{2} - 2 (x - 2)^{2} = 8

x = 5 - 10, x = 5 + 10

Solution steps

(x + 1)^{2} - 2 (x - 2)^{2} = 8

Expand

(x + 1)^{2} - 2 (x - 2)^{2} : - x^{2} + 10 x - 7

- x^{2} + 10 x - 7 = 8

Move

8

to the left side

- x^{2} + 10 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 1 ) ( - 15 )}{2 ( - 1 )}

1 0^{2} - 4 (- 1) (- 15) = 210

x_{1, 2} = \frac{- 10 \pm 2 10}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 10 + 2 10}{2 ( - 1 )}, x_{2} = \frac{- 10 - 2 10}{2 ( - 1 )}

x = \frac{- 10 + 2 10}{2 ( - 1 )} : 5 - 10

x = \frac{- 10 - 2 10}{2 ( - 1 )} : 5 + 10

The solutions to the quadratic equation are:

x = 5 - 10, x = 5 + 10

2 x^{2} + 6 x + 4 = 0

2 x (x + 4) = 42

Q^{2} - 10599.74 Q + 375000 = 0

2 t^{2} + 40 t + 34 = 0

15 x^{2} - 23 x + 4 = 0

What are the solutions to the equation (x+1)^2-2(x-2)^2=8 ?
The solutions to the equation (x+1)^2-2(x-2)^2=8 are x=5-sqrt(10),x=5+sqrt(10)
Find the zeros of (x+1)^2-2(x-2)^2=8
The zeros of (x+1)^2-2(x-2)^2=8 are x=5-sqrt(10),x=5+sqrt(10)