What are the solutions to the equation 3x^2-8x=-1 ?

The solutions to the equation 3x^2-8x=-1 are x=(4+sqrt(13))/3 ,x=(4-sqrt(13))/3

Find the zeros of 3x^2-8x=-1

The zeros of 3x^2-8x=-1 are x=(4+sqrt(13))/3 ,x=(4-sqrt(13))/3

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3 x^{2} - 8 x = - 1

x = \frac{4 + 13}{3}, x = \frac{4 - 13}{3}

Decimal Notation

x = 2.53518 \dots, x = 0.13148 \dots

Solution steps

3 x^{2} - 8 x = - 1

Move

1

to the left side

3 x^{2} - 8 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

(- 8)^{2} - 4 \cdot 3 \cdot 1 = 213

x_{1, 2} = \frac{- ( - 8 ) \pm 2 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 13}{2 \cdot 3}, x_{2} = \frac{- ( - 8 ) - 2 13}{2 \cdot 3}

x = \frac{- ( - 8 ) + 2 13}{2 \cdot 3} : \frac{4 + 13}{3}

x = \frac{- ( - 8 ) - 2 13}{2 \cdot 3} : \frac{4 - 13}{3}

The solutions to the quadratic equation are:

x = \frac{4 + 13}{3}, x = \frac{4 - 13}{3}

4 x + 8 = 2 x + 2

4 ∣ 6 - x ∣ + 2 \geq 10

8 x^{2} + x + 2 = 0

\frac{1}{4} (3)

x 2 = 2

What are the solutions to the equation 3x^2-8x=-1 ?
The solutions to the equation 3x^2-8x=-1 are x=(4+sqrt(13))/3 ,x=(4-sqrt(13))/3
Find the zeros of 3x^2-8x=-1
The zeros of 3x^2-8x=-1 are x=(4+sqrt(13))/3 ,x=(4-sqrt(13))/3