What are the solutions to the equation 4x^2-4x+65=0 ?

The solutions to the equation 4x^2-4x+65=0 are x= 1/2+4i,x= 1/2-4i

Find the zeros of 4x^2-4x+65=0

The zeros of 4x^2-4x+65=0 are x= 1/2+4i,x= 1/2-4i

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4 x^{2} - 4 x + 65 = 0

x = \frac{1}{2} + 4 i, x = \frac{1}{2} - 4 i

Solution steps

4 x^{2} - 4 x + 65 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 \cdot 65}{2 \cdot 4}

Simplify

(- 4)^{2} - 4 \cdot 4 \cdot 65 : 32 i

x_{1, 2} = \frac{- ( - 4 ) \pm 32 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 32 i}{2 \cdot 4}, x_{2} = \frac{- ( - 4 ) - 32 i}{2 \cdot 4}

x = \frac{- ( - 4 ) + 32 i}{2 \cdot 4} : \frac{1}{2} + 4 i

x = \frac{- ( - 4 ) - 32 i}{2 \cdot 4} : \frac{1}{2} - 4 i

The solutions to the quadratic equation are:

x = \frac{1}{2} + 4 i, x = \frac{1}{2} - 4 i

x + 13 = 15

x^{2} - (7 x + 6) = x + 59

∣ x - 2 ∣ = 9

5^{\frac{3}{2}} \cdot 5^{\frac{1}{2}}

[(\frac{1}{3} + \frac{3}{8}) \div (- \frac{3}{5})] + 1

What are the solutions to the equation 4x^2-4x+65=0 ?
The solutions to the equation 4x^2-4x+65=0 are x= 1/2+4i,x= 1/2-4i
Find the zeros of 4x^2-4x+65=0
The zeros of 4x^2-4x+65=0 are x= 1/2+4i,x= 1/2-4i