4.9t^2+17t^2-300=0

Solution for 4.9t^2+17t^2-300=0 equation:

4.9t^2+17t^2-300=0

We add all the numbers together, and all the variables

21.9t^2-300=0

a = 21.9; b = 0; c = -300;
Δ = b2-4ac
Δ = 02-4·21.9·(-300)
Δ = 26280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{26280}=\sqrt{36*730}=\sqrt{36}*\sqrt{730}=6\sqrt{730}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{730}}{2*21.9}=\frac{0-6\sqrt{730}}{43.8}
=-\frac{6\sqrt{730}}{43.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{730}}{2*21.9}=\frac{0+6\sqrt{730}}{43.8}
=\frac{6\sqrt{730}}{43.8}
$