What are the solutions to the equation 40=2x^2+2x ?

The solutions to the equation 40=2x^2+2x are x=4,x=-5

Find the zeros of 40=2x^2+2x

The zeros of 40=2x^2+2x are x=4,x=-5

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40 = 2 x^{2} + 2 x

x = 4, x = - 5

Solution steps

40 = 2 x^{2} + 2 x

Switch sides

2 x^{2} + 2 x = 40

Move

40

to the left side

2 x^{2} + 2 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 40 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 40) = 18

x_{1, 2} = \frac{- 2 \pm 18}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 18}{2 \cdot 2}, x_{2} = \frac{- 2 - 18}{2 \cdot 2}

x = \frac{- 2 + 18}{2 \cdot 2} : 4

x = \frac{- 2 - 18}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = 4, x = - 5

x^{2} - 180 = 0

2 u^{2} - 5 u + 2 = 0

x^{2} - 22 x - 7 = 0

2 x^{2} = 6 + x

8 x - 4 x^{2} = 0

What are the solutions to the equation 40=2x^2+2x ?
The solutions to the equation 40=2x^2+2x are x=4,x=-5
Find the zeros of 40=2x^2+2x
The zeros of 40=2x^2+2x are x=4,x=-5