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

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

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

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

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

x = \frac{1}{2} + i, x = \frac{1}{2} - i

Solution steps

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

Move

5

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 4 \cdot 5 : 8 i

x_{1, 2} = \frac{- ( - 4 ) \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8 i}{2 \cdot 4}, x_{2} = \frac{- ( - 4 ) - 8 i}{2 \cdot 4}

x = \frac{- ( - 4 ) + 8 i}{2 \cdot 4} : \frac{1}{2} + i

x = \frac{- ( - 4 ) - 8 i}{2 \cdot 4} : \frac{1}{2} - i

The solutions to the quadratic equation are:

x = \frac{1}{2} + i, x = \frac{1}{2} - i

x^{2} - 18 x + 81 = 49

2 (x + 10)^{2} - 37 = 13

x^{2} - 72 = 0

x^{2} + x + 2

u, v^{2} = u^{2} + 2 a s

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