What are the solutions to the equation 3(x-6)=5x^2+1 ?

The solutions to the equation 3(x-6)=5x^2+1 are x= 3/10+i(sqrt(371))/(10),x= 3/10-i(sqrt(371))/(10)

Find the zeros of 3(x-6)=5x^2+1

The zeros of 3(x-6)=5x^2+1 are x= 3/10+i(sqrt(371))/(10),x= 3/10-i(sqrt(371))/(10)

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3 (x - 6) = 5 x^{2} + 1

x = \frac{3}{10} + i \frac{371}{10}, x = \frac{3}{10} - i \frac{371}{10}

Solution steps

3 (x - 6) = 5 x^{2} + 1

Expand

3 (x - 6) : 3 x - 18

3 x - 18 = 5 x^{2} + 1

Switch sides

5 x^{2} + 1 = 3 x - 18

Move

18

to the left side

5 x^{2} + 19 = 3 x

Move

3 x

to the left side

5 x^{2} + 19 - 3 x = 0

Write in the standard form

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

5 x^{2} - 3 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 5 \cdot 19}{2 \cdot 5}

Simplify

(- 3)^{2} - 4 \cdot 5 \cdot 19 : 371 i

x_{1, 2} = \frac{- ( - 3 ) \pm 371 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 371 i}{2 \cdot 5}, x_{2} = \frac{- ( - 3 ) - 371 i}{2 \cdot 5}

x = \frac{- ( - 3 ) + 371 i}{2 \cdot 5} : \frac{3}{10} + i \frac{371}{10}

x = \frac{- ( - 3 ) - 371 i}{2 \cdot 5} : \frac{3}{10} - i \frac{371}{10}

The solutions to the quadratic equation are:

x = \frac{3}{10} + i \frac{371}{10}, x = \frac{3}{10} - i \frac{371}{10}

x^{2} - 7 x = - 12

3 x^{2} + 5 x - 28 = 0

- 15 x^{2} - 3.139 x + 11.25 = 0

k^{2} x^{2} + 6 x = (k + 6) x^{2} - k

81 x^{2} - 289 = 0

What are the solutions to the equation 3(x-6)=5x^2+1 ?
The solutions to the equation 3(x-6)=5x^2+1 are x= 3/10+i(sqrt(371))/(10),x= 3/10-i(sqrt(371))/(10)
Find the zeros of 3(x-6)=5x^2+1
The zeros of 3(x-6)=5x^2+1 are x= 3/10+i(sqrt(371))/(10),x= 3/10-i(sqrt(371))/(10)