What are the solutions to the equation (3x+2)(5x-1)=4 ?

The solutions to the equation (3x+2)(5x-1)=4 are x=(-7+sqrt(409))/(30),x=(-7-sqrt(409))/(30)

Find the zeros of (3x+2)(5x-1)=4

The zeros of (3x+2)(5x-1)=4 are x=(-7+sqrt(409))/(30),x=(-7-sqrt(409))/(30)

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(3 x + 2) (5 x - 1) = 4

x = \frac{- 7 + 409}{30}, x = \frac{- 7 - 409}{30}

Decimal Notation

x = 0.44079 \dots, x = - 0.90745 \dots

Solution steps

(3 x + 2) (5 x - 1) = 4

Expand

(3 x + 2) (5 x - 1) : 15 x^{2} + 7 x - 2

15 x^{2} + 7 x - 2 = 4

Move

4

to the left side

15 x^{2} + 7 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 15 ( - 6 )}{2 \cdot 15}

7^{2} - 4 \cdot 15 (- 6) = 409

x_{1, 2} = \frac{- 7 \pm 409}{2 \cdot 15}

Separate the solutions

x_{1} = \frac{- 7 + 409}{2 \cdot 15}, x_{2} = \frac{- 7 - 409}{2 \cdot 15}

x = \frac{- 7 + 409}{2 \cdot 15} : \frac{- 7 + 409}{30}

x = \frac{- 7 - 409}{2 \cdot 15} : \frac{- 7 - 409}{30}

The solutions to the quadratic equation are:

x = \frac{- 7 + 409}{30}, x = \frac{- 7 - 409}{30}

3 t^{2} + 10 t + 1 = t

11 + 12 x^{2} = 10 - 15 x

16.1 x^{2} - 200 x - 200 = 0

x^{2} + 10 x = 2184

(x - 4) (2 - x) + 2 = 1 - x (3 - x)

What are the solutions to the equation (3x+2)(5x-1)=4 ?
The solutions to the equation (3x+2)(5x-1)=4 are x=(-7+sqrt(409))/(30),x=(-7-sqrt(409))/(30)
Find the zeros of (3x+2)(5x-1)=4
The zeros of (3x+2)(5x-1)=4 are x=(-7+sqrt(409))/(30),x=(-7-sqrt(409))/(30)