7+28t-16t^2=0

Solution for 7+28t-16t^2=0 equation:

7+28t-16t^2=0

a = -16; b = 28; c = +7;
Δ = b2-4ac
Δ = 282-4·(-16)·7
Δ = 1232
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:

$\sqrt{\Delta}=\sqrt{1232}=\sqrt{16*77}=\sqrt{16}*\sqrt{77}=4\sqrt{77}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{77}}{2*-16}=\frac{-28-4\sqrt{77}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{77}}{2*-16}=\frac{-28+4\sqrt{77}}{-32}
$