What are the solutions to the equation x^2=7(4/5 x-10) ?

The solutions to the equation x^2=7(4/5 x-10) are x= 14/5+i(sqrt(1554))/5 ,x= 14/5-i(sqrt(1554))/5

Find the zeros of x^2=7(4/5 x-10)

The zeros of x^2=7(4/5 x-10) are x= 14/5+i(sqrt(1554))/5 ,x= 14/5-i(sqrt(1554))/5

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x^{2} = 7 (\frac{4}{5} x - 10)

x = \frac{14}{5} + i \frac{1554}{5}, x = \frac{14}{5} - i \frac{1554}{5}

Solution steps

x^{2} = 7 (\frac{4}{5} x - 10)

Expand

7 (\frac{4}{5} x - 10) : \frac{28}{5} x - 70

x^{2} = \frac{28}{5} x - 70

Move

70

to the left side

x^{2} + 70 = \frac{28}{5} x

Move

\frac{28}{5} x

to the left side

x^{2} + 70 - \frac{28}{5} x = 0

Write in the standard form

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

x^{2} - \frac{28 x}{5} + 70 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - \frac{28}{5} ) \pm ( - \frac{28}{5} ) ^{2} - 4 \cdot 1 \cdot 70}{2 \cdot 1}

Simplify

(- \frac{28}{5})^{2} - 4 \cdot 1 \cdot 70 : i \frac{2 1554}{5}

x_{1, 2} = \frac{- ( - \frac{28}{5} ) \pm i \frac{2 1554}{5}}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - \frac{28}{5} ) + i \frac{2 1554}{5}}{2 \cdot 1}, x_{2} = \frac{- ( - \frac{28}{5} ) - i \frac{2 1554}{5}}{2 \cdot 1}

x = \frac{- ( - \frac{28}{5} ) + i \frac{2 1554}{5}}{2 \cdot 1} : \frac{14}{5} + i \frac{1554}{5}

x = \frac{- ( - \frac{28}{5} ) - i \frac{2 1554}{5}}{2 \cdot 1} : \frac{14}{5} - i \frac{1554}{5}

The solutions to the quadratic equation are:

x = \frac{14}{5} + i \frac{1554}{5}, x = \frac{14}{5} - i \frac{1554}{5}

1 5^{2} + 2 0^{2} = c^{2}

x^{2} - 12 x + 19 = 0

4 x + x^{2} + 16 = - 4 x^{2}

x^{2} - 3.2 x + 1.35 = 0

x, 3 x^{2} - 27 = 0

What are the solutions to the equation x^2=7(4/5 x-10) ?
The solutions to the equation x^2=7(4/5 x-10) are x= 14/5+i(sqrt(1554))/5 ,x= 14/5-i(sqrt(1554))/5
Find the zeros of x^2=7(4/5 x-10)
The zeros of x^2=7(4/5 x-10) are x= 14/5+i(sqrt(1554))/5 ,x= 14/5-i(sqrt(1554))/5