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

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

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

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

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x^{2} - 4 = \frac{5 x}{3}

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

Decimal Notation

x = 3, x = - 1.33333 \dots

Solution steps

x^{2} - 4 = \frac{5 x}{3}

Multiply both sides by

3

3 x^{2} - 12 = 5 x

Move

5 x

to the left side

3 x^{2} - 12 - 5 x = 0

Write in the standard form

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

3 x^{2} - 5 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 12 )}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 (- 12) = 13

x_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 13}{2 \cdot 3}, x_{2} = \frac{- ( - 5 ) - 13}{2 \cdot 3}

x = \frac{- ( - 5 ) + 13}{2 \cdot 3} : 3

x = \frac{- ( - 5 ) - 13}{2 \cdot 3} : - \frac{4}{3}

The solutions to the quadratic equation are:

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

(x + 1)^{3} = x^{3} - 5

(\frac{x - 1}{2})^{2} = 0

x, 4 x^{2} \cdot 4 y^{2} = 12

(x - 3) (x + 1) = (3 x + 1) (2 x + 1)

h, h + s^{2} = h^{2} s

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