What are the solutions to the equation 2x^2+68=24x ?

The solutions to the equation 2x^2+68=24x are x=6+sqrt(2),x=6-sqrt(2)

Find the zeros of 2x^2+68=24x

The zeros of 2x^2+68=24x are x=6+sqrt(2),x=6-sqrt(2)

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2 x^{2} + 68 = 24 x

x = 6 + 2, x = 6 - 2

Solution steps

2 x^{2} + 68 = 24 x

Move

24 x

to the left side

2 x^{2} + 68 - 24 x = 0

Write in the standard form

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

2 x^{2} - 24 x + 68 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 2 \cdot 68}{2 \cdot 2}

(- 24)^{2} - 4 \cdot 2 \cdot 68 = 42

x_{1, 2} = \frac{- ( - 24 ) \pm 4 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 24 ) + 4 2}{2 \cdot 2}, x_{2} = \frac{- ( - 24 ) - 4 2}{2 \cdot 2}

x = \frac{- ( - 24 ) + 4 2}{2 \cdot 2} : 6 + 2

x = \frac{- ( - 24 ) - 4 2}{2 \cdot 2} : 6 - 2

The solutions to the quadratic equation are:

x = 6 + 2, x = 6 - 2

x^{2} + 10 x + 4 = 0

(\frac{4}{6})^{6}

(x + y + z)^{3} - x^{3} - y^{3} - z^{3}

34.46 + \frac{50}{40}

9 x^{2} - 42 x + 49 = 0

What are the solutions to the equation 2x^2+68=24x ?
The solutions to the equation 2x^2+68=24x are x=6+sqrt(2),x=6-sqrt(2)
Find the zeros of 2x^2+68=24x
The zeros of 2x^2+68=24x are x=6+sqrt(2),x=6-sqrt(2)