7=4+4.9*t^2

Solution for 7=4+4.9*t^2 equation:

7=4+4.9t^2

We move all terms to the left:

7-(4+4.9t^2)=0

We get rid of parentheses

-4.9t^2-4+7=0

We add all the numbers together, and all the variables

-4.9t^2+3=0

a = -4.9; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-4.9)·3
Δ = 58.8
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{58.8}}{2*-4.9}=\frac{0-\sqrt{58.8}}{-9.8}
=-\frac{\sqrt{}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{58.8}}{2*-4.9}=\frac{0+\sqrt{58.8}}{-9.8}
=\frac{\sqrt{}}{-9.8}
$