What are the solutions to the equation x^2-12x+29=0 ?

The solutions to the equation x^2-12x+29=0 are x=6+sqrt(7),x=6-sqrt(7)

Find the zeros of x^2-12x+29=0

The zeros of x^2-12x+29=0 are x=6+sqrt(7),x=6-sqrt(7)

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x^{2} - 12 x + 29 = 0

x = 6 + 7, x = 6 - 7

Solution steps

x^{2} - 12 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 29 = 27

x_{1, 2} = \frac{- ( - 12 ) \pm 2 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 7}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 2 7}{2 \cdot 1}

x = \frac{- ( - 12 ) + 2 7}{2 \cdot 1} : 6 + 7

x = \frac{- ( - 12 ) - 2 7}{2 \cdot 1} : 6 - 7

The solutions to the quadratic equation are:

x = 6 + 7, x = 6 - 7

- 15 m^{2} - 73 m + 10

r^{2} + 2 r + 2 = 0

\frac{3 π}{2} \cdot 3

2^{3^{2}}

x^{2} - 4 + 4 = 0

What are the solutions to the equation x^2-12x+29=0 ?
The solutions to the equation x^2-12x+29=0 are x=6+sqrt(7),x=6-sqrt(7)
Find the zeros of x^2-12x+29=0
The zeros of x^2-12x+29=0 are x=6+sqrt(7),x=6-sqrt(7)