What are the solutions to the equation s^2-13s+36=0 ?

The solutions to the equation s^2-13s+36=0 are s=9,s=4

Find the zeros of s^2-13s+36=0

The zeros of s^2-13s+36=0 are s=9,s=4

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s^{2} - 13 s + 36 = 0

s = 9, s = 4

Solution steps

s^{2} - 13 s + 36 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 36 = 5

s_{1, 2} = \frac{- ( - 13 ) \pm 5}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- ( - 13 ) + 5}{2 \cdot 1}, s_{2} = \frac{- ( - 13 ) - 5}{2 \cdot 1}

s = \frac{- ( - 13 ) + 5}{2 \cdot 1} : 9

s = \frac{- ( - 13 ) - 5}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

s = 9, s = 4

x^{2} = - 11 x - 24

2 x^{2} + 4 x - 10

4905 x^{2} - 12 x - 70 = 0

3 x^{2} - 12 x - 5 = 0

(x - 8) (x - 2) = - 9

What are the solutions to the equation s^2-13s+36=0 ?
The solutions to the equation s^2-13s+36=0 are s=9,s=4
Find the zeros of s^2-13s+36=0
The zeros of s^2-13s+36=0 are s=9,s=4