t^2=148/4.9

Solution for t^2=148/4.9 equation:

t^2=148/4.9

We move all terms to the left:

t^2-(148/4.9)=0

We add all the numbers together, and all the variables

t^2-(+148/4.9)=0

We get rid of parentheses

t^2-148/4.9=0

We multiply all the terms by the denominator


t^2*4.9-148=0

Wy multiply elements

4.9t^2-148=0

a = 4.9; b = 0; c = -148;
Δ = b2-4ac
Δ = 02-4·4.9·(-148)
Δ = 2900.8
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{2900.8}}{2*4.9}=\frac{0-\sqrt{2900.8}}{9.8}
=-\frac{\sqrt{}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{2900.8}}{2*4.9}=\frac{0+\sqrt{2900.8}}{9.8}
=\frac{\sqrt{}}{9.8}
$