What are the solutions to the equation 2x^2+142=-36x ?

The solutions to the equation 2x^2+142=-36x are x=-9+sqrt(10),x=-9-sqrt(10)

Find the zeros of 2x^2+142=-36x

The zeros of 2x^2+142=-36x are x=-9+sqrt(10),x=-9-sqrt(10)

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2 x^{2} + 142 = - 36 x

x = - 9 + 10, x = - 9 - 10

Decimal Notation

x = - 5.83772 \dots, x = - 12.16227 \dots

Solution steps

2 x^{2} + 142 = - 36 x

Move

36 x

to the left side

2 x^{2} + 142 + 36 x = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} + 36 x + 142 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 2 \cdot 142}{2 \cdot 2}

3 6^{2} - 4 \cdot 2 \cdot 142 = 410

x_{1, 2} = \frac{- 36 \pm 4 10}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 36 + 4 10}{2 \cdot 2}, x_{2} = \frac{- 36 - 4 10}{2 \cdot 2}

x = \frac{- 36 + 4 10}{2 \cdot 2} : - 9 + 10

x = \frac{- 36 - 4 10}{2 \cdot 2} : - 9 - 10

The solutions to the quadratic equation are:

x = - 9 + 10, x = - 9 - 10

3 + 6 y - 2 y^{2} = 0

x (\frac{x}{2}) = 800

x^{2} - 7 x + 6

16 - x^{2} = (3 x - 4)^{2}

n^{2} + 3 n + 2 = 0

What are the solutions to the equation 2x^2+142=-36x ?
The solutions to the equation 2x^2+142=-36x are x=-9+sqrt(10),x=-9-sqrt(10)
Find the zeros of 2x^2+142=-36x
The zeros of 2x^2+142=-36x are x=-9+sqrt(10),x=-9-sqrt(10)