What are the solutions to the equation (x+4)^2=2x(5x-1)-7(x-2) ?

The solutions to the equation (x+4)^2=2x(5x-1)-7(x-2) are x=-1/9 ,x=2

Find the zeros of (x+4)^2=2x(5x-1)-7(x-2)

The zeros of (x+4)^2=2x(5x-1)-7(x-2) are x=-1/9 ,x=2

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(x + 4)^{2} = 2 x (5 x - 1) - 7 (x - 2)

x = - \frac{1}{9}, x = 2

Decimal Notation

x = - 0.11111 \dots, x = 2

Solution steps

(x + 4)^{2} = 2 x (5 x - 1) - 7 (x - 2)

Expand

(x + 4)^{2} : x^{2} + 8 x + 16

Expand

2 x (5 x - 1) - 7 (x - 2) : 10 x^{2} - 9 x + 14

x^{2} + 8 x + 16 = 10 x^{2} - 9 x + 14

Move

14

to the left side

x^{2} + 8 x + 2 = 10 x^{2} - 9 x

Move

9 x

to the left side

x^{2} + 17 x + 2 = 10 x^{2}

Move

10 x^{2}

to the left side

- 9 x^{2} + 17 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 ( - 9 ) \cdot 2}{2 ( - 9 )}

1 7^{2} - 4 (- 9) \cdot 2 = 19

x_{1, 2} = \frac{- 17 \pm 19}{2 ( - 9 )}

Separate the solutions

x_{1} = \frac{- 17 + 19}{2 ( - 9 )}, x_{2} = \frac{- 17 - 19}{2 ( - 9 )}

x = \frac{- 17 + 19}{2 ( - 9 )} : - \frac{1}{9}

x = \frac{- 17 - 19}{2 ( - 9 )} : 2

The solutions to the quadratic equation are:

x = - \frac{1}{9}, x = 2

ln (ab)

\frac{11}{42} + \frac{( - 20 )}{42}

125 m^{4} - 20 n^{4}

- \frac{1}{\frac{1}{4}}

\frac{1}{3 ^{5}}

What are the solutions to the equation (x+4)^2=2x(5x-1)-7(x-2) ?
The solutions to the equation (x+4)^2=2x(5x-1)-7(x-2) are x=-1/9 ,x=2
Find the zeros of (x+4)^2=2x(5x-1)-7(x-2)
The zeros of (x+4)^2=2x(5x-1)-7(x-2) are x=-1/9 ,x=2