What are the solutions to the equation roots x^2-12x+52 ?

The solutions to the equation roots x^2-12x+52 are x=6+4i,x=6-4i

Find the zeros of roots x^2-12x+52

The zeros of roots x^2-12x+52 are x=6+4i,x=6-4i

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roots

x^{2} - 12 x + 52

x = 6 + 4 i, x = 6 - 4 i

Solution steps

x^{2} - 12 x + 52

The roots are the intercepts with the x-axis

(y = 0)

x^{2} - 12 x + 52 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 52}{2 \cdot 1}

Simplify

(- 12)^{2} - 4 \cdot 1 \cdot 52 : 8 i

x_{1, 2} = \frac{- ( - 12 ) \pm 8 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 8 i}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 8 i}{2 \cdot 1}

x = \frac{- ( - 12 ) + 8 i}{2 \cdot 1} : 6 + 4 i

x = \frac{- ( - 12 ) - 8 i}{2 \cdot 1} : 6 - 4 i

The solutions to the quadratic equation are:

x = 6 + 4 i, x = 6 - 4 i

2 x^{2} + 1

7 x^{2} + 36 x - 1440 = 0

y, x^{2} + 14 x y + 49 y^{2} = 100

4 x (2 x - 1) = 9 - 7 x

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

What are the solutions to the equation roots x^2-12x+52 ?
The solutions to the equation roots x^2-12x+52 are x=6+4i,x=6-4i
Find the zeros of roots x^2-12x+52
The zeros of roots x^2-12x+52 are x=6+4i,x=6-4i