What are the solutions to the equation 2(x+3)^2-5=14x+17 ?

The solutions to the equation 2(x+3)^2-5=14x+17 are x=2,x=-1

Find the zeros of 2(x+3)^2-5=14x+17

The zeros of 2(x+3)^2-5=14x+17 are x=2,x=-1

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2 (x + 3)^{2} - 5 = 14 x + 17

x = 2, x = - 1

Solution steps

2 (x + 3)^{2} - 5 = 14 x + 17

Expand

2 (x + 3)^{2} - 5 : 2 x^{2} + 12 x + 13

2 x^{2} + 12 x + 13 = 14 x + 17

Move

17

to the left side

2 x^{2} + 12 x - 4 = 14 x

Move

14 x

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 4) = 6

x_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 2}

x = \frac{- ( - 2 ) + 6}{2 \cdot 2} : 2

x = \frac{- ( - 2 ) - 6}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = 2, x = - 1

10 x^{2} + 8 x = 0

(x + 4)^{2} - 2 = 7

2 y (5 y + 6) = 7 y

25 (x + 2)^{2} = (x - 7)^{2} - 81

k^{2} - 4 k + 1 = - 19

What are the solutions to the equation 2(x+3)^2-5=14x+17 ?
The solutions to the equation 2(x+3)^2-5=14x+17 are x=2,x=-1
Find the zeros of 2(x+3)^2-5=14x+17
The zeros of 2(x+3)^2-5=14x+17 are x=2,x=-1