5(t)=-4.9t^2+1.5t+17

Solution for 5(t)=-4.9t^2+1.5t+17 equation:

5(t)=-4.9t^2+1.5t+17

We move all terms to the left:

5(t)-(-4.9t^2+1.5t+17)=0

We get rid of parentheses

4.9t^2-1.5t+5t-17=0

We add all the numbers together, and all the variables

4.9t^2+3.5t-17=0

a = 4.9; b = 3.5; c = -17;
Δ = b2-4ac
Δ = 3.52-4·4.9·(-17)
Δ = 345.45
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3.5)-\sqrt{345.45}}{2*4.9}=\frac{-3.5-\sqrt{345.45}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{345.45}}{2*4.9}=\frac{-3.5+\sqrt{345.45}}{9.8}
$