What are the solutions to the equation 12x^2+x+7=5x^2+5x ?

The solutions to the equation 12x^2+x+7=5x^2+5x are x= 2/7+i(3sqrt(5))/7 ,x= 2/7-i(3sqrt(5))/7

Find the zeros of 12x^2+x+7=5x^2+5x

The zeros of 12x^2+x+7=5x^2+5x are x= 2/7+i(3sqrt(5))/7 ,x= 2/7-i(3sqrt(5))/7

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12 x^{2} + x + 7 = 5 x^{2} + 5 x

x = \frac{2}{7} + i \frac{3 5}{7}, x = \frac{2}{7} - i \frac{3 5}{7}

Solution steps

12 x^{2} + x + 7 = 5 x^{2} + 5 x

Move

5 x

to the left side

12 x^{2} - 4 x + 7 = 5 x^{2}

Move

5 x^{2}

to the left side

7 x^{2} - 4 x + 7 = 0

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 7 \cdot 7 : 65 i

x_{1, 2} = \frac{- ( - 4 ) \pm 6 5 i}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 6 5 i}{2 \cdot 7}, x_{2} = \frac{- ( - 4 ) - 6 5 i}{2 \cdot 7}

x = \frac{- ( - 4 ) + 6 5 i}{2 \cdot 7} : \frac{2}{7} + i \frac{3 5}{7}

x = \frac{- ( - 4 ) - 6 5 i}{2 \cdot 7} : \frac{2}{7} - i \frac{3 5}{7}

The solutions to the quadratic equation are:

x = \frac{2}{7} + i \frac{3 5}{7}, x = \frac{2}{7} - i \frac{3 5}{7}

(x + 1) (2 x - 4) = 6 x

x^{2} + 8 x + 23 = 0

r^{2} + 10 r - 45 = - 6

x, x^{2} + y^{2} - 6 x + 14 y + 62 = 0

x^{2} = - 10 x

What are the solutions to the equation 12x^2+x+7=5x^2+5x ?
The solutions to the equation 12x^2+x+7=5x^2+5x are x= 2/7+i(3sqrt(5))/7 ,x= 2/7-i(3sqrt(5))/7
Find the zeros of 12x^2+x+7=5x^2+5x
The zeros of 12x^2+x+7=5x^2+5x are x= 2/7+i(3sqrt(5))/7 ,x= 2/7-i(3sqrt(5))/7