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

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

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

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

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

b = 6

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

b = 6

The solution to the quadratic equation is:

b = 6

(x + 1) (2 x + 5) = (2 x + 3) (x - 4) + 5

\frac{5}{2} - 5

4 x + 6 = 38

4 x - 2 > 8 x - 2

x = (x - 6)^{2}

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