What is the answer to (3x-7)^2-5(2x+1)(x-2)=-x^2-[-(3x+1)] ?

The answer to (3x-7)^2-5(2x+1)(x-2)=-x^2-[-(3x+1)] is x= 29/15

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(3x-7)^2-5(2x+1)(x-2)=-x^2-[-(3x+1)]

(3 x - 7)^{2} - 5 (2 x + 1) (x - 2) = - x^{2} - [- (3 x + 1)]

x = \frac{29}{15}

Decimal Notation

x = 1.93333 \dots

Solution steps

(3 x - 7)^{2} - 5 (2 x + 1) (x - 2) = - x^{2} - [- (3 x + 1)]

(3 x - 7)^{2} - 5 (2 x + 1) (x - 2) = - x^{2} + 3 x + 1

Expand

(3 x - 7)^{2} - 5 (2 x + 1) (x - 2) : - x^{2} - 27 x + 59

- x^{2} - 27 x + 59 = - x^{2} + 3 x + 1

Move

59

to the right side

- x^{2} - 27 x = - x^{2} + 3 x - 58

Subtract

- x^{2} + 3 x

from both sides

- x^{2} - 27 x - (- x^{2} + 3 x) = - x^{2} + 3 x - 58 - (- x^{2} + 3 x)

Simplify

- 30 x = - 58

Divide both sides by

- 30

x = \frac{29}{15}

- 2.6 b + 4 = 0.9 b - 17

- 20 = \frac{z}{10} - 22

x, 2 x - y = 12

\frac{2 x + 3}{4} = \frac{x + 7}{3}

r + 6 = - 9 - 9 - 2 r

What is the answer to (3x-7)^2-5(2x+1)(x-2)=-x^2-[-(3x+1)] ?
The answer to (3x-7)^2-5(2x+1)(x-2)=-x^2-[-(3x+1)] is x= 29/15
Solve for x: (3x-7)^2-5(2x+1)(x-2)=-x^2-[-(3x+1)]
The solution is x= 29/15