What are the solutions to the equation 3x^2-4x=5 ?

The solutions to the equation 3x^2-4x=5 are x=(2+sqrt(19))/3 ,x=(2-sqrt(19))/3

Find the zeros of 3x^2-4x=5

The zeros of 3x^2-4x=5 are x=(2+sqrt(19))/3 ,x=(2-sqrt(19))/3

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3 x^{2} - 4 x = 5

x = \frac{2 + 19}{3}, x = \frac{2 - 19}{3}

Solution steps

3 x^{2} - 4 x = 5

Move

5

to the left side

3 x^{2} - 4 x - 5 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 3 (- 5) = 219

x_{1, 2} = \frac{- ( - 4 ) \pm 2 19}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 19}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2 19}{2 \cdot 3}

x = \frac{- ( - 4 ) + 2 19}{2 \cdot 3} : \frac{2 + 19}{3}

x = \frac{- ( - 4 ) - 2 19}{2 \cdot 3} : \frac{2 - 19}{3}

The solutions to the quadratic equation are:

x = \frac{2 + 19}{3}, x = \frac{2 - 19}{3}

x^{2} - 4 = 32

4 x (x - 3) = - 9

x^{2} - 4 = 12

(x + 3)^{2} = 10

(x + 3)^{2} = 20

What are the solutions to the equation 3x^2-4x=5 ?
The solutions to the equation 3x^2-4x=5 are x=(2+sqrt(19))/3 ,x=(2-sqrt(19))/3
Find the zeros of 3x^2-4x=5
The zeros of 3x^2-4x=5 are x=(2+sqrt(19))/3 ,x=(2-sqrt(19))/3