What are the solutions to the equation ((2x+1)^2)/5-((x+3)(x-3))/3 = 20/3 ?

The solutions to the equation ((2x+1)^2)/5-((x+3)(x-3))/3 = 20/3 are x=2,x=-26/7

Find the zeros of ((2x+1)^2)/5-((x+3)(x-3))/3 = 20/3

The zeros of ((2x+1)^2)/5-((x+3)(x-3))/3 = 20/3 are x=2,x=-26/7

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\frac{( 2 x + 1 ) ^{2}}{5} - \frac{( x + 3 ) ( x - 3 )}{3} = \frac{20}{3}

x = 2, x = - \frac{26}{7}

Decimal Notation

x = 2, x = - 3.71428 \dots

Solution steps

\frac{( 2 x + 1 ) ^{2}}{5} - \frac{( x + 3 ) ( x - 3 )}{3} = \frac{20}{3}

Find Least Common Multiplier of

5, 3 : 15

Multiply by LCM=

15

\frac{( 2 x + 1 ) ^{2}}{5} \cdot 15 - \frac{( x + 3 ) ( x - 3 )}{3} \cdot 15 = \frac{20}{3} \cdot 15

Simplify

3 (2 x + 1)^{2} - 5 (x + 3) (x - 3) = 100

Expand

3 (2 x + 1)^{2} - 5 (x + 3) (x - 3) : 7 x^{2} + 12 x + 48

7 x^{2} + 12 x + 48 = 100

Move

100

to the left side

7 x^{2} + 12 x - 52 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 7 ( - 52 )}{2 \cdot 7}

1 2^{2} - 4 \cdot 7 (- 52) = 40

x_{1, 2} = \frac{- 12 \pm 40}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 12 + 40}{2 \cdot 7}, x_{2} = \frac{- 12 - 40}{2 \cdot 7}

x = \frac{- 12 + 40}{2 \cdot 7} : 2

x = \frac{- 12 - 40}{2 \cdot 7} : - \frac{26}{7}

The solutions to the quadratic equation are:

x = 2, x = - \frac{26}{7}

0.1 - 0.01 (x + 2) = - 0.03 (4 - x)

(- 3 x^{3} y^{4})^{2}

\frac{7 b ^{4}}{4 b ^{- 1}}

\frac{17 x ^{0} y ^{- 8}}{4 ^{- 2} z ^{- 6}}

2 x^{2} - 4 x - 70

What are the solutions to the equation ((2x+1)^2)/5-((x+3)(x-3))/3 = 20/3 ?
The solutions to the equation ((2x+1)^2)/5-((x+3)(x-3))/3 = 20/3 are x=2,x=-26/7
Find the zeros of ((2x+1)^2)/5-((x+3)(x-3))/3 = 20/3
The zeros of ((2x+1)^2)/5-((x+3)(x-3))/3 = 20/3 are x=2,x=-26/7