What are the solutions to the equation -4x^2+5x=3 ?

The solutions to the equation -4x^2+5x=3 are x= 5/8-i(sqrt(23))/8 ,x= 5/8+i(sqrt(23))/8

Find the zeros of -4x^2+5x=3

The zeros of -4x^2+5x=3 are x= 5/8-i(sqrt(23))/8 ,x= 5/8+i(sqrt(23))/8

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- 4 x^{2} + 5 x = 3

x = \frac{5}{8} - i \frac{23}{8}, x = \frac{5}{8} + i \frac{23}{8}

Solution steps

- 4 x^{2} + 5 x = 3

Move

3

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

Simplify

5^{2} - 4 (- 4) (- 3) : 23 i

x_{1, 2} = \frac{- 5 \pm 23 i}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 5 + 23 i}{2 ( - 4 )}, x_{2} = \frac{- 5 - 23 i}{2 ( - 4 )}

x = \frac{- 5 + 23 i}{2 ( - 4 )} : \frac{5}{8} - i \frac{23}{8}

x = \frac{- 5 - 23 i}{2 ( - 4 )} : \frac{5}{8} + i \frac{23}{8}

The solutions to the quadratic equation are:

x = \frac{5}{8} - i \frac{23}{8}, x = \frac{5}{8} + i \frac{23}{8}

7 = b^{2} - 6 b

5 a^{2} - 5

2 x + 6 = x^{2} - 3 x

b^{4} (\frac{1}{3} b^{2}) (12 b^{- 8})

∣ x ∣ = 49

What are the solutions to the equation -4x^2+5x=3 ?
The solutions to the equation -4x^2+5x=3 are x= 5/8-i(sqrt(23))/8 ,x= 5/8+i(sqrt(23))/8
Find the zeros of -4x^2+5x=3
The zeros of -4x^2+5x=3 are x= 5/8-i(sqrt(23))/8 ,x= 5/8+i(sqrt(23))/8