What are the solutions to the equation 20=8t+(2t^2)/2 ?

The solutions to the equation 20=8t+(2t^2)/2 are t=2,t=-10

Find the zeros of 20=8t+(2t^2)/2

The zeros of 20=8t+(2t^2)/2 are t=2,t=-10

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

20 = 8 t + \frac{2 t ^{2}}{2}

t = 2, t = - 10

Solution steps

20 = 8 t + \frac{2 t ^{2}}{2}

Multiply both sides by

2

40 = 16 t + 2 t^{2}

Switch sides

16 t + 2 t^{2} = 40

Move

40

to the left side

16 t + 2 t^{2} - 40 = 0

Write in the standard form

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

2 t^{2} + 16 t - 40 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 2 ( - 40 )}{2 \cdot 2}

1 6^{2} - 4 \cdot 2 (- 40) = 24

t_{1, 2} = \frac{- 16 \pm 24}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 16 + 24}{2 \cdot 2}, t_{2} = \frac{- 16 - 24}{2 \cdot 2}

t = \frac{- 16 + 24}{2 \cdot 2} : 2

t = \frac{- 16 - 24}{2 \cdot 2} : - 10

The solutions to the quadratic equation are:

t = 2, t = - 10

a^{\frac{1}{2}} \cdot a^{\frac{1}{2}}

3 x^{2} - 10 x + 7

8 x^{2} + 16 x - 68 = 3

\frac{3}{4} x = 12

9 (y - 5) = - 9

What are the solutions to the equation 20=8t+(2t^2)/2 ?
The solutions to the equation 20=8t+(2t^2)/2 are t=2,t=-10
Find the zeros of 20=8t+(2t^2)/2
The zeros of 20=8t+(2t^2)/2 are t=2,t=-10