What are the solutions to the equation 121x^2-462x+441=360-72x ?

The solutions to the equation 121x^2-462x+441=360-72x are x=3,x= 27/121

Find the zeros of 121x^2-462x+441=360-72x

The zeros of 121x^2-462x+441=360-72x are x=3,x= 27/121

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121 x^{2} - 462 x + 441 = 360 - 72 x

x = 3, x = \frac{27}{121}

Decimal Notation

x = 3, x = 0.22314 \dots

Solution steps

121 x^{2} - 462 x + 441 = 360 - 72 x

Move

72 x

to the left side

121 x^{2} - 390 x + 441 = 360

Move

360

to the left side

121 x^{2} - 390 x + 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 390 ) \pm ( - 390 ) ^{2} - 4 \cdot 121 \cdot 81}{2 \cdot 121}

(- 390)^{2} - 4 \cdot 121 \cdot 81 = 336

x_{1, 2} = \frac{- ( - 390 ) \pm 336}{2 \cdot 121}

Separate the solutions

x_{1} = \frac{- ( - 390 ) + 336}{2 \cdot 121}, x_{2} = \frac{- ( - 390 ) - 336}{2 \cdot 121}

x = \frac{- ( - 390 ) + 336}{2 \cdot 121} : 3

x = \frac{- ( - 390 ) - 336}{2 \cdot 121} : \frac{27}{121}

The solutions to the quadratic equation are:

x = 3, x = \frac{27}{121}

- 10 x^{2} + 180 = 0

6 x^{2} - 10 x - 19 = 0

x, f (x + h) = 3 x^{2} + 4 x - 2

x, (y - x - 2)^{2} = 9 x

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

What are the solutions to the equation 121x^2-462x+441=360-72x ?
The solutions to the equation 121x^2-462x+441=360-72x are x=3,x= 27/121
Find the zeros of 121x^2-462x+441=360-72x
The zeros of 121x^2-462x+441=360-72x are x=3,x= 27/121