What are the solutions to the equation 180=(x+20)x128 ?

The solutions to the equation 180=(x+20)x128 are x=(-80+sqrt(6490))/8 ,x=-(80+sqrt(6490))/8

Find the zeros of 180=(x+20)x128

The zeros of 180=(x+20)x128 are x=(-80+sqrt(6490))/8 ,x=-(80+sqrt(6490))/8

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180 = (x + 20) x 128

x = \frac{- 80 + 6490}{8}, x = - \frac{80 + 6490}{8}

Decimal Notation

x = 0.07006 \dots, x = - 20.07006 \dots

Solution steps

180 = (x + 20) x \cdot 128

Expand

(x + 20) x \cdot 128 : 128 x^{2} + 2560 x

180 = 128 x^{2} + 2560 x

Switch sides

128 x^{2} + 2560 x = 180

Move

180

to the left side

128 x^{2} + 2560 x - 180 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2560 \pm 256 0 ^{2} - 4 \cdot 128 ( - 180 )}{2 \cdot 128}

256 0^{2} - 4 \cdot 128 (- 180) = 326490

x_{1, 2} = \frac{- 2560 \pm 32 6490}{2 \cdot 128}

Separate the solutions

x_{1} = \frac{- 2560 + 32 6490}{2 \cdot 128}, x_{2} = \frac{- 2560 - 32 6490}{2 \cdot 128}

x = \frac{- 2560 + 32 6490}{2 \cdot 128} : \frac{- 80 + 6490}{8}

x = \frac{- 2560 - 32 6490}{2 \cdot 128} : - \frac{80 + 6490}{8}

The solutions to the quadratic equation are:

x = \frac{- 80 + 6490}{8}, x = - \frac{80 + 6490}{8}

y, x^{2} - 4 y^{2} = 0

y, x^{2} + 5 x + y^{2} - y + 7 = 0

x, y = 2 - 2 x^{2}

2 x^{2} - 8 = 120

x^{2} - a

What are the solutions to the equation 180=(x+20)x128 ?
The solutions to the equation 180=(x+20)x128 are x=(-80+sqrt(6490))/8 ,x=-(80+sqrt(6490))/8
Find the zeros of 180=(x+20)x128
The zeros of 180=(x+20)x128 are x=(-80+sqrt(6490))/8 ,x=-(80+sqrt(6490))/8