What are the solutions to the equation 3x^2-9=7x ?

The solutions to the equation 3x^2-9=7x are x=(7+sqrt(157))/6 ,x=(7-sqrt(157))/6

Find the zeros of 3x^2-9=7x

The zeros of 3x^2-9=7x are x=(7+sqrt(157))/6 ,x=(7-sqrt(157))/6

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3 x^{2} - 9 = 7 x

x = \frac{7 + 157}{6}, x = \frac{7 - 157}{6}

Solution steps

3 x^{2} - 9 = 7 x

Move

7 x

to the left side

3 x^{2} - 9 - 7 x = 0

Write in the standard form

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

3 x^{2} - 7 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 3 ( - 9 )}{2 \cdot 3}

(- 7)^{2} - 4 \cdot 3 (- 9) = 157

x_{1, 2} = \frac{- ( - 7 ) \pm 157}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 157}{2 \cdot 3}, x_{2} = \frac{- ( - 7 ) - 157}{2 \cdot 3}

x = \frac{- ( - 7 ) + 157}{2 \cdot 3} : \frac{7 + 157}{6}

x = \frac{- ( - 7 ) - 157}{2 \cdot 3} : \frac{7 - 157}{6}

The solutions to the quadratic equation are:

x = \frac{7 + 157}{6}, x = \frac{7 - 157}{6}

3 x - 2 + 2 x + 5 = 3 (x + 5)

x^{2} - 7 x + 6 = 0

lo g_{49} (343)

\frac{1}{2} x + 3 < 4 x - 7

What are the solutions to the equation 3x^2-9=7x ?
The solutions to the equation 3x^2-9=7x are x=(7+sqrt(157))/6 ,x=(7-sqrt(157))/6
Find the zeros of 3x^2-9=7x
The zeros of 3x^2-9=7x are x=(7+sqrt(157))/6 ,x=(7-sqrt(157))/6