What are the solutions to the equation 2x^2+22x+36=0 ?

The solutions to the equation 2x^2+22x+36=0 are x=-2,x=-9

Find the zeros of 2x^2+22x+36=0

The zeros of 2x^2+22x+36=0 are x=-2,x=-9

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2 x^{2} + 22 x + 36 = 0

x = - 2, x = - 9

Solution steps

2 x^{2} + 22 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 \cdot 2 \cdot 36}{2 \cdot 2}

2 2^{2} - 4 \cdot 2 \cdot 36 = 14

x_{1, 2} = \frac{- 22 \pm 14}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 22 + 14}{2 \cdot 2}, x_{2} = \frac{- 22 - 14}{2 \cdot 2}

x = \frac{- 22 + 14}{2 \cdot 2} : - 2

x = \frac{- 22 - 14}{2 \cdot 2} : - 9

The solutions to the quadratic equation are:

x = - 2, x = - 9

y^{2} + 8 y + 11 = 0

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

3^{2} + y^{2} = 5^{2}

x^{2} + 6 x = 3

2 x^{2} + 9 x + 14 = 0

What are the solutions to the equation 2x^2+22x+36=0 ?
The solutions to the equation 2x^2+22x+36=0 are x=-2,x=-9
Find the zeros of 2x^2+22x+36=0
The zeros of 2x^2+22x+36=0 are x=-2,x=-9