What are the solutions to the equation 3x^2=7x-3 ?

The solutions to the equation 3x^2=7x-3 are x=(7+sqrt(13))/6 ,x=(7-sqrt(13))/6

Find the zeros of 3x^2=7x-3

The zeros of 3x^2=7x-3 are x=(7+sqrt(13))/6 ,x=(7-sqrt(13))/6

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3 x^{2} = 7 x - 3

x = \frac{7 + 13}{6}, x = \frac{7 - 13}{6}

Decimal Notation

x = 1.76759 \dots, x = 0.56574 \dots

Solution steps

3 x^{2} = 7 x - 3

Move

3

to the left side

3 x^{2} + 3 = 7 x

Move

7 x

to the left side

3 x^{2} + 3 - 7 x = 0

Write in the standard form

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

3 x^{2} - 7 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

(- 7)^{2} - 4 \cdot 3 \cdot 3 = 13

x_{1, 2} = \frac{- ( - 7 ) \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 13}{2 \cdot 3}, x_{2} = \frac{- ( - 7 ) - 13}{2 \cdot 3}

x = \frac{- ( - 7 ) + 13}{2 \cdot 3} : \frac{7 + 13}{6}

x = \frac{- ( - 7 ) - 13}{2 \cdot 3} : \frac{7 - 13}{6}

The solutions to the quadratic equation are:

x = \frac{7 + 13}{6}, x = \frac{7 - 13}{6}

2 x - 4 (x + 5) \leq 2 (x + 2)

2 x + 6 = 9 x + 8

\frac{8 + 2}{7 + 7}

10 h^{2} - 9 h - 9

91

What are the solutions to the equation 3x^2=7x-3 ?
The solutions to the equation 3x^2=7x-3 are x=(7+sqrt(13))/6 ,x=(7-sqrt(13))/6
Find the zeros of 3x^2=7x-3
The zeros of 3x^2=7x-3 are x=(7+sqrt(13))/6 ,x=(7-sqrt(13))/6