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

The solutions to the equation (x-5)(3x+1)=-1 are x=(7+sqrt(61))/3 ,x=(7-sqrt(61))/3

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

The zeros of (x-5)(3x+1)=-1 are x=(7+sqrt(61))/3 ,x=(7-sqrt(61))/3

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

x = \frac{7 + 61}{3}, x = \frac{7 - 61}{3}

Decimal Notation

x = 4.93674 \dots, x = - 0.27008 \dots

Solution steps

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

Expand

(x - 5) (3 x + 1) : 3 x^{2} - 14 x - 5

3 x^{2} - 14 x - 5 = - 1

Move

1

to the left side

3 x^{2} - 14 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

(- 14)^{2} - 4 \cdot 3 (- 4) = 261

x_{1, 2} = \frac{- ( - 14 ) \pm 2 61}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2 61}{2 \cdot 3}, x_{2} = \frac{- ( - 14 ) - 2 61}{2 \cdot 3}

x = \frac{- ( - 14 ) + 2 61}{2 \cdot 3} : \frac{7 + 61}{3}

x = \frac{- ( - 14 ) - 2 61}{2 \cdot 3} : \frac{7 - 61}{3}

The solutions to the quadratic equation are:

x = \frac{7 + 61}{3}, x = \frac{7 - 61}{3}

x^{2} + (2 a - 1) x - 2 a = 0

z (z - 1) = 3

4 m^{2} + 10 m + 5 = - m

3 x^{2} + 5 = 4 x^{2} + 3

y^{2} - 2 y + 17 = 0

What are the solutions to the equation (x-5)(3x+1)=-1 ?
The solutions to the equation (x-5)(3x+1)=-1 are x=(7+sqrt(61))/3 ,x=(7-sqrt(61))/3
Find the zeros of (x-5)(3x+1)=-1
The zeros of (x-5)(3x+1)=-1 are x=(7+sqrt(61))/3 ,x=(7-sqrt(61))/3