What are the solutions to the equation 0=x^2-12x+36 ?

The solutions to the equation 0=x^2-12x+36 are x=6

Find the zeros of 0=x^2-12x+36

The zeros of 0=x^2-12x+36 are x=6

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0 = x^{2} - 12 x + 36

x = 6

Solution steps

0 = x^{2} - 12 x + 36

Switch sides

x^{2} - 12 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 36 = 0

x_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 1}

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

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

x = 6

The solution to the quadratic equation is:

x = 6

3 x^{2} - 6 x - 13 = 0

1 5^{2} = 9^{2} + x^{2}

(x + 7)^{2} + 7 (x + 7) + 6 = 0

2 x^{2} - 5 x + 14 = 0

y^{2} = - 25

What are the solutions to the equation 0=x^2-12x+36 ?
The solutions to the equation 0=x^2-12x+36 are x=6
Find the zeros of 0=x^2-12x+36
The zeros of 0=x^2-12x+36 are x=6