What are the solutions to the equation (3x-8)(x+1)=3 ?

The solutions to the equation (3x-8)(x+1)=3 are x=(5+sqrt(157))/6 ,x=(5-sqrt(157))/6

Find the zeros of (3x-8)(x+1)=3

The zeros of (3x-8)(x+1)=3 are x=(5+sqrt(157))/6 ,x=(5-sqrt(157))/6

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(3 x - 8) (x + 1) = 3

x = \frac{5 + 157}{6}, x = \frac{5 - 157}{6}

Decimal Notation

x = 2.92166 \dots, x = - 1.25499 \dots

Solution steps

(3 x - 8) (x + 1) = 3

Expand

(3 x - 8) (x + 1) : 3 x^{2} - 5 x - 8

3 x^{2} - 5 x - 8 = 3

Move

3

to the left side

3 x^{2} - 5 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 11 )}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 (- 11) = 157

x_{1, 2} = \frac{- ( - 5 ) \pm 157}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 157}{2 \cdot 3}, x_{2} = \frac{- ( - 5 ) - 157}{2 \cdot 3}

x = \frac{- ( - 5 ) + 157}{2 \cdot 3} : \frac{5 + 157}{6}

x = \frac{- ( - 5 ) - 157}{2 \cdot 3} : \frac{5 - 157}{6}

The solutions to the quadratic equation are:

x = \frac{5 + 157}{6}, x = \frac{5 - 157}{6}

3 x^{2} - 6 x - 3 = 0

- z = 2

∣ x - 2 ∣ = ∣ 3 x + 5 ∣

\frac{1 0 ^{7}}{1 0 ^{4}}

9 \cdot a^{3} - 4 \cdot a

What are the solutions to the equation (3x-8)(x+1)=3 ?
The solutions to the equation (3x-8)(x+1)=3 are x=(5+sqrt(157))/6 ,x=(5-sqrt(157))/6
Find the zeros of (3x-8)(x+1)=3
The zeros of (3x-8)(x+1)=3 are x=(5+sqrt(157))/6 ,x=(5-sqrt(157))/6