-4.9t^2-140.2+150=0

Solution for -4.9t^2-140.2+150=0 equation:

-4.9t^2-140.2+150=0

We add all the numbers together, and all the variables

-4.9t^2+9.8=0

a = -4.9; b = 0; c = +9.8;
Δ = b2-4ac
Δ = 02-4·(-4.9)·9.8
Δ = 192.08
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{-(0)-\sqrt{192.08}}{2*-4.9}=\frac{0-\sqrt{192.08}}{-9.8}
=-\frac{\sqrt{}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{192.08}}{2*-4.9}=\frac{0+\sqrt{192.08}}{-9.8}
=\frac{\sqrt{}}{-9.8}
$