What are the solutions to the equation 3x^2+5x=3 ?

The solutions to the equation 3x^2+5x=3 are x=(-5+sqrt(61))/6 ,x=(-5-sqrt(61))/6

Find the zeros of 3x^2+5x=3

The zeros of 3x^2+5x=3 are x=(-5+sqrt(61))/6 ,x=(-5-sqrt(61))/6

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3 x^{2} + 5 x = 3

x = \frac{- 5 + 61}{6}, x = \frac{- 5 - 61}{6}

Decimal Notation

x = 0.46837 \dots, x = - 2.13504 \dots

Solution steps

3 x^{2} + 5 x = 3

Move

3

to the left side

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 3 (- 3) = 61

x_{1, 2} = \frac{- 5 \pm 61}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 5 + 61}{2 \cdot 3}, x_{2} = \frac{- 5 - 61}{2 \cdot 3}

x = \frac{- 5 + 61}{2 \cdot 3} : \frac{- 5 + 61}{6}

x = \frac{- 5 - 61}{2 \cdot 3} : \frac{- 5 - 61}{6}

The solutions to the quadratic equation are:

x = \frac{- 5 + 61}{6}, x = \frac{- 5 - 61}{6}

3 x = 2 x + 2

(- \frac{6}{L ^{2}} + \frac{12 x}{L ^{3}})^{2}

\frac{26}{1 + 5 i}

\frac{1}{4} x - \frac{1}{5} (x - 1) = 5 x

4 x^{2} - 2 = 0

What are the solutions to the equation 3x^2+5x=3 ?
The solutions to the equation 3x^2+5x=3 are x=(-5+sqrt(61))/6 ,x=(-5-sqrt(61))/6
Find the zeros of 3x^2+5x=3
The zeros of 3x^2+5x=3 are x=(-5+sqrt(61))/6 ,x=(-5-sqrt(61))/6