What are the solutions to the equation 180=0.05s^2+1.1s ?

The solutions to the equation 180=0.05s^2+1.1s are s=50,s=-72

Find the zeros of 180=0.05s^2+1.1s

The zeros of 180=0.05s^2+1.1s are s=50,s=-72

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180 = 0.05 s^{2} + 1.1 s

s = 50, s = - 72

Solution steps

180 = 0.05 s^{2} + 1.1 s

Multiply both sides by

100

18000 = 5 s^{2} + 110 s

Switch sides

5 s^{2} + 110 s = 18000

Move

18000

to the left side

5 s^{2} + 110 s - 18000 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 110 \pm 11 0 ^{2} - 4 \cdot 5 ( - 18000 )}{2 \cdot 5}

11 0^{2} - 4 \cdot 5 (- 18000) = 610

s_{1, 2} = \frac{- 110 \pm 610}{2 \cdot 5}

Separate the solutions

s_{1} = \frac{- 110 + 610}{2 \cdot 5}, s_{2} = \frac{- 110 - 610}{2 \cdot 5}

s = \frac{- 110 + 610}{2 \cdot 5} : 50

s = \frac{- 110 - 610}{2 \cdot 5} : - 72

The solutions to the quadratic equation are:

s = 50, s = - 72

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

2 x^{2} - 5 x - 6

x^{2} + 5 x - 36 = 0

2 (x - 2)^{2} - 39 = - 31

x^{2} + 8 x - 24 = 3 x

What are the solutions to the equation 180=0.05s^2+1.1s ?
The solutions to the equation 180=0.05s^2+1.1s are s=50,s=-72
Find the zeros of 180=0.05s^2+1.1s
The zeros of 180=0.05s^2+1.1s are s=50,s=-72