What are the solutions to the equation -4x^2+16x=19 ?

The solutions to the equation -4x^2+16x=19 are x=2-i(sqrt(3))/2 ,x=2+i(sqrt(3))/2

Find the zeros of -4x^2+16x=19

The zeros of -4x^2+16x=19 are x=2-i(sqrt(3))/2 ,x=2+i(sqrt(3))/2

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- 4 x^{2} + 16 x = 19

x = 2 - i \frac{3}{2}, x = 2 + i \frac{3}{2}

Solution steps

- 4 x^{2} + 16 x = 19

Move

19

to the left side

- 4 x^{2} + 16 x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 4 ) ( - 19 )}{2 ( - 4 )}

Simplify

1 6^{2} - 4 (- 4) (- 19) : 43 i

x_{1, 2} = \frac{- 16 \pm 4 3 i}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 16 + 4 3 i}{2 ( - 4 )}, x_{2} = \frac{- 16 - 4 3 i}{2 ( - 4 )}

x = \frac{- 16 + 4 3 i}{2 ( - 4 )} : 2 - i \frac{3}{2}

x = \frac{- 16 - 4 3 i}{2 ( - 4 )} : 2 + i \frac{3}{2}

The solutions to the quadratic equation are:

x = 2 - i \frac{3}{2}, x = 2 + i \frac{3}{2}

30 r^{2} - 19 r - 5 = 0

18 x^{2} - 6 x + 1 = 0

5 + \frac{3}{2}

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

\frac{3 ( x + 1 )}{6} = 2 x - 1

What are the solutions to the equation -4x^2+16x=19 ?
The solutions to the equation -4x^2+16x=19 are x=2-i(sqrt(3))/2 ,x=2+i(sqrt(3))/2
Find the zeros of -4x^2+16x=19
The zeros of -4x^2+16x=19 are x=2-i(sqrt(3))/2 ,x=2+i(sqrt(3))/2