What are the solutions to the equation 7x^2+3=8x ?

The solutions to the equation 7x^2+3=8x are x= 4/7+i(sqrt(5))/7 ,x= 4/7-i(sqrt(5))/7

Find the zeros of 7x^2+3=8x

The zeros of 7x^2+3=8x are x= 4/7+i(sqrt(5))/7 ,x= 4/7-i(sqrt(5))/7

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7 x^{2} + 3 = 8 x

x = \frac{4}{7} + i \frac{5}{7}, x = \frac{4}{7} - i \frac{5}{7}

Solution steps

7 x^{2} + 3 = 8 x

Move

8 x

to the left side

7 x^{2} + 3 - 8 x = 0

Write in the standard form

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

7 x^{2} - 8 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 7 \cdot 3}{2 \cdot 7}

Simplify

(- 8)^{2} - 4 \cdot 7 \cdot 3 : 25 i

x_{1, 2} = \frac{- ( - 8 ) \pm 2 5 i}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 5 i}{2 \cdot 7}, x_{2} = \frac{- ( - 8 ) - 2 5 i}{2 \cdot 7}

x = \frac{- ( - 8 ) + 2 5 i}{2 \cdot 7} : \frac{4}{7} + i \frac{5}{7}

x = \frac{- ( - 8 ) - 2 5 i}{2 \cdot 7} : \frac{4}{7} - i \frac{5}{7}

The solutions to the quadratic equation are:

x = \frac{4}{7} + i \frac{5}{7}, x = \frac{4}{7} - i \frac{5}{7}

x^{2} - 1 > - x^{2} + 1

(x - 1) (x + 2) < 0

(\frac{x}{11})^{- 2}

3 y + 12 = - 6

2 (6 + 2 x) = 8 x

What are the solutions to the equation 7x^2+3=8x ?
The solutions to the equation 7x^2+3=8x are x= 4/7+i(sqrt(5))/7 ,x= 4/7-i(sqrt(5))/7
Find the zeros of 7x^2+3=8x
The zeros of 7x^2+3=8x are x= 4/7+i(sqrt(5))/7 ,x= 4/7-i(sqrt(5))/7