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

The solutions to the equation 4x^2-2x=-5 are x= 1/4+i(sqrt(19))/4 ,x= 1/4-i(sqrt(19))/4

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

The zeros of 4x^2-2x=-5 are x= 1/4+i(sqrt(19))/4 ,x= 1/4-i(sqrt(19))/4

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

x = \frac{1}{4} + i \frac{19}{4}, x = \frac{1}{4} - i \frac{19}{4}

Solution steps

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

Move

5

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 4 \cdot 5 : 219 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 19 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 19 i}{2 \cdot 4}, x_{2} = \frac{- ( - 2 ) - 2 19 i}{2 \cdot 4}

x = \frac{- ( - 2 ) + 2 19 i}{2 \cdot 4} : \frac{1}{4} + i \frac{19}{4}

x = \frac{- ( - 2 ) - 2 19 i}{2 \cdot 4} : \frac{1}{4} - i \frac{19}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{19}{4}, x = \frac{1}{4} - i \frac{19}{4}

x^{2} + 5 = 10

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

x^{2} + 5 = 54

x^{2} + 4 x - 240 = 0

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

What are the solutions to the equation 4x^2-2x=-5 ?
The solutions to the equation 4x^2-2x=-5 are x= 1/4+i(sqrt(19))/4 ,x= 1/4-i(sqrt(19))/4
Find the zeros of 4x^2-2x=-5
The zeros of 4x^2-2x=-5 are x= 1/4+i(sqrt(19))/4 ,x= 1/4-i(sqrt(19))/4