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

The solutions to the equation 2(3x+5)+2(x-5)=2(3x+5)(x-5) are x=(7+2sqrt(31))/3 ,x=(7-2sqrt(31))/3

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

The zeros of 2(3x+5)+2(x-5)=2(3x+5)(x-5) are x=(7+2sqrt(31))/3 ,x=(7-2sqrt(31))/3

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2 (3 x + 5) + 2 (x - 5) = 2 (3 x + 5) (x - 5)

x = \frac{7 + 2 31}{3}, x = \frac{7 - 2 31}{3}

Decimal Notation

x = 6.04517 \dots, x = - 1.37850 \dots

Solution steps

2 (3 x + 5) + 2 (x - 5) = 2 (3 x + 5) (x - 5)

Expand

2 (3 x + 5) + 2 (x - 5) : 8 x

Expand

2 (3 x + 5) (x - 5) : 6 x^{2} - 20 x - 50

8 x = 6 x^{2} - 20 x - 50

Switch sides

6 x^{2} - 20 x - 50 = 8 x

Move

8 x

to the left side

6 x^{2} - 28 x - 50 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 6 ( - 50 )}{2 \cdot 6}

(- 28)^{2} - 4 \cdot 6 (- 50) = 831

x_{1, 2} = \frac{- ( - 28 ) \pm 8 31}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 8 31}{2 \cdot 6}, x_{2} = \frac{- ( - 28 ) - 8 31}{2 \cdot 6}

x = \frac{- ( - 28 ) + 8 31}{2 \cdot 6} : \frac{7 + 2 31}{3}

x = \frac{- ( - 28 ) - 8 31}{2 \cdot 6} : \frac{7 - 2 31}{3}

The solutions to the quadratic equation are:

x = \frac{7 + 2 31}{3}, x = \frac{7 - 2 31}{3}

x^{2} + 9 x + 15 = 0

2 5^{2} - 7^{2} = x^{2}

X^{2} - 5 X + 4 = 0

L^{2} - 24 L + 32 = 0

28090 = 25000 (1 + x)^{2}

What are the solutions to the equation 2(3x+5)+2(x-5)=2(3x+5)(x-5) ?
The solutions to the equation 2(3x+5)+2(x-5)=2(3x+5)(x-5) are x=(7+2sqrt(31))/3 ,x=(7-2sqrt(31))/3
Find the zeros of 2(3x+5)+2(x-5)=2(3x+5)(x-5)
The zeros of 2(3x+5)+2(x-5)=2(3x+5)(x-5) are x=(7+2sqrt(31))/3 ,x=(7-2sqrt(31))/3