What are the solutions to the equation 5x^2+4=7x ?

The solutions to the equation 5x^2+4=7x are x= 7/10+i(sqrt(31))/(10),x= 7/10-i(sqrt(31))/(10)

Find the zeros of 5x^2+4=7x

The zeros of 5x^2+4=7x are x= 7/10+i(sqrt(31))/(10),x= 7/10-i(sqrt(31))/(10)

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5 x^{2} + 4 = 7 x

x = \frac{7}{10} + i \frac{31}{10}, x = \frac{7}{10} - i \frac{31}{10}

Solution steps

5 x^{2} + 4 = 7 x

Move

7 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 7)^{2} - 4 \cdot 5 \cdot 4 : 31 i

x_{1, 2} = \frac{- ( - 7 ) \pm 31 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 31 i}{2 \cdot 5}, x_{2} = \frac{- ( - 7 ) - 31 i}{2 \cdot 5}

x = \frac{- ( - 7 ) + 31 i}{2 \cdot 5} : \frac{7}{10} + i \frac{31}{10}

x = \frac{- ( - 7 ) - 31 i}{2 \cdot 5} : \frac{7}{10} - i \frac{31}{10}

The solutions to the quadratic equation are:

x = \frac{7}{10} + i \frac{31}{10}, x = \frac{7}{10} - i \frac{31}{10}

p (p - 3) 2 - 4 (p - 3) 3 = 0

a, x^{2} + a^{2} y^{2} = 4 a^{2}

3 x^{2} + 24 x + 45 = 0

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

3 x^{2} - 4 x = 15

What are the solutions to the equation 5x^2+4=7x ?
The solutions to the equation 5x^2+4=7x are x= 7/10+i(sqrt(31))/(10),x= 7/10-i(sqrt(31))/(10)
Find the zeros of 5x^2+4=7x
The zeros of 5x^2+4=7x are x= 7/10+i(sqrt(31))/(10),x= 7/10-i(sqrt(31))/(10)