What are the solutions to the equation 44=1.5(2/3)v+0.06v^2 ?

The solutions to the equation 44=1.5(2/3)v+0.06v^2 are v=20,v=-110/3

Find the zeros of 44=1.5(2/3)v+0.06v^2

The zeros of 44=1.5(2/3)v+0.06v^2 are v=20,v=-110/3

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44 = 1.5 (\frac{2}{3}) v + 0.06 v^{2}

v = 20, v = - \frac{110}{3}

Solution steps

44 = 1.5 (\frac{2}{3}) v + 0.06 v^{2}

Multiply both sides by

100

4400 = 100 v + 6 v^{2}

Switch sides

100 v + 6 v^{2} = 4400

Move

4400

to the left side

100 v + 6 v^{2} - 4400 = 0

Write in the standard form

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

6 v^{2} + 100 v - 4400 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 \cdot 6 ( - 4400 )}{2 \cdot 6}

10 0^{2} - 4 \cdot 6 (- 4400) = 340

v_{1, 2} = \frac{- 100 \pm 340}{2 \cdot 6}

Separate the solutions

v_{1} = \frac{- 100 + 340}{2 \cdot 6}, v_{2} = \frac{- 100 - 340}{2 \cdot 6}

v = \frac{- 100 + 340}{2 \cdot 6} : 20

v = \frac{- 100 - 340}{2 \cdot 6} : - \frac{110}{3}

The solutions to the quadratic equation are:

v = 20, v = - \frac{110}{3}

m + 1 = - 4 m^{2}

1 6^{2} = 1 2^{2} + x^{2}

8 x - 10 = x^{2} - 2 x

x^{2} + 10 - 24 = 0

2 x^{2} - 8 x + 24 = 0

What are the solutions to the equation 44=1.5(2/3)v+0.06v^2 ?
The solutions to the equation 44=1.5(2/3)v+0.06v^2 are v=20,v=-110/3
Find the zeros of 44=1.5(2/3)v+0.06v^2
The zeros of 44=1.5(2/3)v+0.06v^2 are v=20,v=-110/3