What are the solutions to the equation 4x^2-20x+29=0 ?

The solutions to the equation 4x^2-20x+29=0 are x= 5/2+i,x= 5/2-i

Find the zeros of 4x^2-20x+29=0

The zeros of 4x^2-20x+29=0 are x= 5/2+i,x= 5/2-i

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4 x^{2} - 20 x + 29 = 0

x = \frac{5}{2} + i, x = \frac{5}{2} - i

Solution steps

4 x^{2} - 20 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 29}{2 \cdot 4}

Simplify

(- 20)^{2} - 4 \cdot 4 \cdot 29 : 8 i

x_{1, 2} = \frac{- ( - 20 ) \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 8 i}{2 \cdot 4}, x_{2} = \frac{- ( - 20 ) - 8 i}{2 \cdot 4}

x = \frac{- ( - 20 ) + 8 i}{2 \cdot 4} : \frac{5}{2} + i

x = \frac{- ( - 20 ) - 8 i}{2 \cdot 4} : \frac{5}{2} - i

The solutions to the quadratic equation are:

x = \frac{5}{2} + i, x = \frac{5}{2} - i

(x + 5)^{2} + 5 (2 x + 3) = 2 (x + 4)^{2}

4 x^{2} + 40 x + 100 = 0

x^{2} - 8 x + 96 = 0

x, 3 x^{2} - 5 y^{6} = 2 x

2 (3 x + 1)^{2} - 8 = 0

What are the solutions to the equation 4x^2-20x+29=0 ?
The solutions to the equation 4x^2-20x+29=0 are x= 5/2+i,x= 5/2-i
Find the zeros of 4x^2-20x+29=0
The zeros of 4x^2-20x+29=0 are x= 5/2+i,x= 5/2-i