What are the solutions to the equation 2(9x^2+6x+1)=(3x+1)(x-2) ?

The solutions to the equation 2(9x^2+6x+1)=(3x+1)(x-2) are x=-1/3 ,x=-4/5

Find the zeros of 2(9x^2+6x+1)=(3x+1)(x-2)

The zeros of 2(9x^2+6x+1)=(3x+1)(x-2) are x=-1/3 ,x=-4/5

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2 (9 x^{2} + 6 x + 1) = (3 x + 1) (x - 2)

x = - \frac{1}{3}, x = - \frac{4}{5}

Decimal

x = - 0.33333 \dots, x = - 0.8

Solution steps

2 (9 x^{2} + 6 x + 1) = (3 x + 1) (x - 2)

Move

(3 x + 1) (x - 2)

to the left side

2 (9 x^{2} + 6 x + 1) - (3 x + 1) (x - 2) = 0

Factor

2 (9 x^{2} + 6 x + 1) - (3 x + 1) (x - 2) : (3 x + 1) (5 x + 4)

(3 x + 1) (5 x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x + 1 = 0 or 5 x + 4 = 0

Solve

3 x + 1 = 0 : x = - \frac{1}{3}

Solve

5 x + 4 = 0 : x = - \frac{4}{5}

The solutions to the quadratic equation are:

x = - \frac{1}{3}, x = - \frac{4}{5}

64 - x^{2}

2 a^{2} + 8 a - 15 = 3 a - 3

100 = r^{2} + r^{2} - 2 \cdot r \cdot r \cdot \frac{3}{2}

x (3 x + 4) = 5

21 x^{2} - 16 x - 16

What are the solutions to the equation 2(9x^2+6x+1)=(3x+1)(x-2) ?
The solutions to the equation 2(9x^2+6x+1)=(3x+1)(x-2) are x=-1/3 ,x=-4/5
Find the zeros of 2(9x^2+6x+1)=(3x+1)(x-2)
The zeros of 2(9x^2+6x+1)=(3x+1)(x-2) are x=-1/3 ,x=-4/5