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

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

Find the zeros of u^2+13u+36=0

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

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u^{2} + 13 u + 36 = 0

u = - 4, u = - 9

Solution steps

u^{2} + 13 u + 36 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 36 = 5

u_{1, 2} = \frac{- 13 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 13 + 5}{2 \cdot 1}, u_{2} = \frac{- 13 - 5}{2 \cdot 1}

u = \frac{- 13 + 5}{2 \cdot 1} : - 4

u = \frac{- 13 - 5}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

u = - 4, u = - 9

- 2 x + 1 = - x^{2} + 4

4 - x^{2} = 2 + x

x^{2} + \frac{x}{2} = 5

6 x^{2} - 8 x = 3 x - 4

4 x^{2} + 3 x = 24 - x

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