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4.9t^2+19.8t-20=0? How do i find t?

 Jan 29, 2015

Best Answer 

 #2
avatar+26364 
+5

4.9t^2+19.8t-20=0? How do i find t ?

$$\small{\text{
Use the formula:
$
\boxed{
ax^2+bx+c=0 \qquad x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}
}
$
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{19.8^2-4*4.9*(-20)} }{2*4.9}
$
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{ 392.04+392 } }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{ 784.04 } }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm 28.001 }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1}=\dfrac{-19.8 + 28.001 }{9.8} = 0.8368
$
}}
}}\\\\
\small{\text{
$
t_{2}=\dfrac{-19.8 - 28.001 }{9.8} = -4.8777
$
}}$$

 Jan 30, 2015
 #1
avatar+118587 
+5

I'd use the quadratic formula  

 Jan 30, 2015
 #2
avatar+26364 
+5
Best Answer

4.9t^2+19.8t-20=0? How do i find t ?

$$\small{\text{
Use the formula:
$
\boxed{
ax^2+bx+c=0 \qquad x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}
}
$
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{19.8^2-4*4.9*(-20)} }{2*4.9}
$
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{ 392.04+392 } }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{ 784.04 } }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm 28.001 }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1}=\dfrac{-19.8 + 28.001 }{9.8} = 0.8368
$
}}
}}\\\\
\small{\text{
$
t_{2}=\dfrac{-19.8 - 28.001 }{9.8} = -4.8777
$
}}$$

heureka Jan 30, 2015

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