What are the solutions to the equation 3x^2+243=60x ?

The solutions to the equation 3x^2+243=60x are x=10+sqrt(19),x=10-sqrt(19)

Find the zeros of 3x^2+243=60x

The zeros of 3x^2+243=60x are x=10+sqrt(19),x=10-sqrt(19)

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3 x^{2} + 243 = 60 x

x = 10 + 19, x = 10 - 19

Decimal Notation

x = 14.35889 \dots, x = 5.64110 \dots

Solution steps

3 x^{2} + 243 = 60 x

Move

60 x

to the left side

3 x^{2} + 243 - 60 x = 0

Write in the standard form

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

3 x^{2} - 60 x + 243 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 60 ) \pm ( - 60 ) ^{2} - 4 \cdot 3 \cdot 243}{2 \cdot 3}

(- 60)^{2} - 4 \cdot 3 \cdot 243 = 619

x_{1, 2} = \frac{- ( - 60 ) \pm 6 19}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 60 ) + 6 19}{2 \cdot 3}, x_{2} = \frac{- ( - 60 ) - 6 19}{2 \cdot 3}

x = \frac{- ( - 60 ) + 6 19}{2 \cdot 3} : 10 + 19

x = \frac{- ( - 60 ) - 6 19}{2 \cdot 3} : 10 - 19

The solutions to the quadratic equation are:

x = 10 + 19, x = 10 - 19

46 = - 16 t^{2} + 50 t + 10

2 n x^{2} = x + \frac{1}{n}

- x^{2} - 5 x - 4 = 0

x^{2} - 10 x = 10

x^{2} + 8 x + 12 = 0

What are the solutions to the equation 3x^2+243=60x ?
The solutions to the equation 3x^2+243=60x are x=10+sqrt(19),x=10-sqrt(19)
Find the zeros of 3x^2+243=60x
The zeros of 3x^2+243=60x are x=10+sqrt(19),x=10-sqrt(19)