What are the solutions to the equation 3x^2-1=5x ?

The solutions to the equation 3x^2-1=5x are x=(5+sqrt(37))/6 ,x=(5-sqrt(37))/6

Find the zeros of 3x^2-1=5x

The zeros of 3x^2-1=5x are x=(5+sqrt(37))/6 ,x=(5-sqrt(37))/6

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3 x^{2} - 1 = 5 x

x = \frac{5 + 37}{6}, x = \frac{5 - 37}{6}

Solution steps

3 x^{2} - 1 = 5 x

Move

5 x

to the left side

3 x^{2} - 1 - 5 x = 0

Write in the standard form

a x^{2} + b x + c = 0

3 x^{2} - 5 x - 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 3 (- 1) = 37

x_{1, 2} = \frac{- ( - 5 ) \pm 37}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 37}{2 \cdot 3}, x_{2} = \frac{- ( - 5 ) - 37}{2 \cdot 3}

x = \frac{- ( - 5 ) + 37}{2 \cdot 3} : \frac{5 + 37}{6}

x = \frac{- ( - 5 ) - 37}{2 \cdot 3} : \frac{5 - 37}{6}

The solutions to the quadratic equation are:

x = \frac{5 + 37}{6}, x = \frac{5 - 37}{6}

\frac{2.688 \cdot 1 0 ^{2}}{4.8 \cdot 1 0 ^{- 3}}

\frac{12}{5}

2 k + 3 (k + 4) = - 3

\frac{7}{3} (6 x + 6) \leq 6 x - 6

(4 t - 6)^{2} - 12 (4 t - 6) + 20 = 0

What are the solutions to the equation 3x^2-1=5x ?
The solutions to the equation 3x^2-1=5x are x=(5+sqrt(37))/6 ,x=(5-sqrt(37))/6
Find the zeros of 3x^2-1=5x
The zeros of 3x^2-1=5x are x=(5+sqrt(37))/6 ,x=(5-sqrt(37))/6