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

The solutions to the equation 4x^2+4=5x are x= 5/8+i(sqrt(39))/8 ,x= 5/8-i(sqrt(39))/8

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

The zeros of 4x^2+4=5x are x= 5/8+i(sqrt(39))/8 ,x= 5/8-i(sqrt(39))/8

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4 x^{2} + 4 = 5 x

x = \frac{5}{8} + i \frac{39}{8}, x = \frac{5}{8} - i \frac{39}{8}

Solution steps

4 x^{2} + 4 = 5 x

Move

5 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 4 \cdot 4 : 39 i

x_{1, 2} = \frac{- ( - 5 ) \pm 39 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 39 i}{2 \cdot 4}, x_{2} = \frac{- ( - 5 ) - 39 i}{2 \cdot 4}

x = \frac{- ( - 5 ) + 39 i}{2 \cdot 4} : \frac{5}{8} + i \frac{39}{8}

x = \frac{- ( - 5 ) - 39 i}{2 \cdot 4} : \frac{5}{8} - i \frac{39}{8}

The solutions to the quadratic equation are:

x = \frac{5}{8} + i \frac{39}{8}, x = \frac{5}{8} - i \frac{39}{8}

- 0.0625 x^{2} + 2 x + 50 = 60

x, f (x - 3) = x^{2} - 1

x, f (x - 3) = x^{2} - 2

x^{2} - 529 = 0

2 x + 4 = (x - 2)^{2}

What are the solutions to the equation 4x^2+4=5x ?
The solutions to the equation 4x^2+4=5x are x= 5/8+i(sqrt(39))/8 ,x= 5/8-i(sqrt(39))/8
Find the zeros of 4x^2+4=5x
The zeros of 4x^2+4=5x are x= 5/8+i(sqrt(39))/8 ,x= 5/8-i(sqrt(39))/8