What are the solutions to the equation 3x^2-4x+9=10x+3 ?

The solutions to the equation 3x^2-4x+9=10x+3 are x=(7+sqrt(31))/3 ,x=(7-sqrt(31))/3

Find the zeros of 3x^2-4x+9=10x+3

The zeros of 3x^2-4x+9=10x+3 are x=(7+sqrt(31))/3 ,x=(7-sqrt(31))/3

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3 x^{2} - 4 x + 9 = 10 x + 3

x = \frac{7 + 31}{3}, x = \frac{7 - 31}{3}

Decimal Notation

x = 4.18925 \dots, x = 0.47741 \dots

Solution steps

3 x^{2} - 4 x + 9 = 10 x + 3

Move

3

to the left side

3 x^{2} - 4 x + 6 = 10 x

Move

10 x

to the left side

3 x^{2} - 14 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

(- 14)^{2} - 4 \cdot 3 \cdot 6 = 231

x_{1, 2} = \frac{- ( - 14 ) \pm 2 31}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2 31}{2 \cdot 3}, x_{2} = \frac{- ( - 14 ) - 2 31}{2 \cdot 3}

x = \frac{- ( - 14 ) + 2 31}{2 \cdot 3} : \frac{7 + 31}{3}

x = \frac{- ( - 14 ) - 2 31}{2 \cdot 3} : \frac{7 - 31}{3}

The solutions to the quadratic equation are:

x = \frac{7 + 31}{3}, x = \frac{7 - 31}{3}

x^{2} - 14 x = 6

y, 2 x^{2} - 3 y^{2} + 5 x y - 2 x + 3 y - 4 = 0

4 x^{2} + 7 mn x - 15 m^{2} n^{2}

x, z = x^{2} - 3 y^{2} + 5 x y

y, 6 x^{2} y - 9 x^{2} + 23 x y - 6 x + 21 y + 80 = 0

What are the solutions to the equation 3x^2-4x+9=10x+3 ?
The solutions to the equation 3x^2-4x+9=10x+3 are x=(7+sqrt(31))/3 ,x=(7-sqrt(31))/3
Find the zeros of 3x^2-4x+9=10x+3
The zeros of 3x^2-4x+9=10x+3 are x=(7+sqrt(31))/3 ,x=(7-sqrt(31))/3