What are the solutions to the equation (2x+5)(6x-1)=120 ?

The solutions to the equation (2x+5)(6x-1)=120 are x=(-7+2sqrt(106))/6 ,x=-(7+2sqrt(106))/6

Find the zeros of (2x+5)(6x-1)=120

The zeros of (2x+5)(6x-1)=120 are x=(-7+2sqrt(106))/6 ,x=-(7+2sqrt(106))/6

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(2 x + 5) (6 x - 1) = 120

x = \frac{- 7 + 2 106}{6}, x = - \frac{7 + 2 106}{6}

Decimal Notation

x = 2.26521 \dots, x = - 4.59854 \dots

Solution steps

(2 x + 5) (6 x - 1) = 120

Expand

(2 x + 5) (6 x - 1) : 12 x^{2} + 28 x - 5

12 x^{2} + 28 x - 5 = 120

Move

120

to the left side

12 x^{2} + 28 x - 125 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 12 ( - 125 )}{2 \cdot 12}

2 8^{2} - 4 \cdot 12 (- 125) = 8106

x_{1, 2} = \frac{- 28 \pm 8 106}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 28 + 8 106}{2 \cdot 12}, x_{2} = \frac{- 28 - 8 106}{2 \cdot 12}

x = \frac{- 28 + 8 106}{2 \cdot 12} : \frac{- 7 + 2 106}{6}

x = \frac{- 28 - 8 106}{2 \cdot 12} : - \frac{7 + 2 106}{6}

The solutions to the quadratic equation are:

x = \frac{- 7 + 2 106}{6}, x = - \frac{7 + 2 106}{6}

24 x^{2} - 82 x + 48 = 0

5 y^{2} = 80

- 0.5 t^{2} + 3.2 t + 22 = 24

1010 x^{2} + 1009 x - 1008 = 0

- 7 = 2 x^{2} - 5

What are the solutions to the equation (2x+5)(6x-1)=120 ?
The solutions to the equation (2x+5)(6x-1)=120 are x=(-7+2sqrt(106))/6 ,x=-(7+2sqrt(106))/6
Find the zeros of (2x+5)(6x-1)=120
The zeros of (2x+5)(6x-1)=120 are x=(-7+2sqrt(106))/6 ,x=-(7+2sqrt(106))/6