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

Solution for 0=17(t)+-4.9t^2 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We use the square of the difference formula

-(17t-4.9t^2)=0

We get rid of parentheses

4.9t^2-17t=0

a = 4.9; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·4.9·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*4.9}=\frac{0}{9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*4.9}=\frac{34}{9.8}
=3+2.875/6.125
$