What are the solutions to the equation 4x^2-4x=-9 ?

The solutions to the equation 4x^2-4x=-9 are x= 1/2+sqrt(2)i,x= 1/2-sqrt(2)i

Find the zeros of 4x^2-4x=-9

The zeros of 4x^2-4x=-9 are x= 1/2+sqrt(2)i,x= 1/2-sqrt(2)i

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4 x^{2} - 4 x = - 9

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

Solution steps

4 x^{2} - 4 x = - 9

Move

9

to the left side

4 x^{2} - 4 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 \cdot 9}{2 \cdot 4}

Simplify

(- 4)^{2} - 4 \cdot 4 \cdot 9 : 82 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

100 v^{2} - 182 v + 82.56 = 0

\frac{x ^{\frac{2}{3}}}{x ^{2}}

x^{3} (y^{- 5}) (x^{- 8})

∣ x + 2 ∣ < ∣ 2 x ∣

6 x - 8 = 16 + 10 x

What are the solutions to the equation 4x^2-4x=-9 ?
The solutions to the equation 4x^2-4x=-9 are x= 1/2+sqrt(2)i,x= 1/2-sqrt(2)i
Find the zeros of 4x^2-4x=-9
The zeros of 4x^2-4x=-9 are x= 1/2+sqrt(2)i,x= 1/2-sqrt(2)i