What are the solutions to the equation 3x^2+6=10x ?

The solutions to the equation 3x^2+6=10x are x=(5+sqrt(7))/3 ,x=(5-sqrt(7))/3

Find the zeros of 3x^2+6=10x

The zeros of 3x^2+6=10x are x=(5+sqrt(7))/3 ,x=(5-sqrt(7))/3

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3 x^{2} + 6 = 10 x

x = \frac{5 + 7}{3}, x = \frac{5 - 7}{3}

Decimal Notation

x = 2.54858 \dots, x = 0.78474 \dots

Solution steps

3 x^{2} + 6 = 10 x

Move

10 x

to the left side

3 x^{2} + 6 - 10 x = 0

Write in the standard form

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

3 x^{2} - 10 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

(- 10)^{2} - 4 \cdot 3 \cdot 6 = 27

x_{1, 2} = \frac{- ( - 10 ) \pm 2 7}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 7}{2 \cdot 3}, x_{2} = \frac{- ( - 10 ) - 2 7}{2 \cdot 3}

x = \frac{- ( - 10 ) + 2 7}{2 \cdot 3} : \frac{5 + 7}{3}

x = \frac{- ( - 10 ) - 2 7}{2 \cdot 3} : \frac{5 - 7}{3}

The solutions to the quadratic equation are:

x = \frac{5 + 7}{3}, x = \frac{5 - 7}{3}

3 y \geq 9

2 \leq 3 - x \leq 12

n + 8 \leq 10

25 x^{2} y^{3}

X^{2} + 3 X - 10 = 0

What are the solutions to the equation 3x^2+6=10x ?
The solutions to the equation 3x^2+6=10x are x=(5+sqrt(7))/3 ,x=(5-sqrt(7))/3
Find the zeros of 3x^2+6=10x
The zeros of 3x^2+6=10x are x=(5+sqrt(7))/3 ,x=(5-sqrt(7))/3