4.9t(t-4)-103=0

Solution for 4.9t(t-4)-103=0 equation:

4.9t(t-4)-103=0

We multiply parentheses

4t^2-16t-103=0

a = 4; b = -16; c = -103;
Δ = b2-4ac
Δ = -162-4·4·(-103)
Δ = 1904
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:

$\sqrt{\Delta}=\sqrt{1904}=\sqrt{16*119}=\sqrt{16}*\sqrt{119}=4\sqrt{119}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{119}}{2*4}=\frac{16-4\sqrt{119}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{119}}{2*4}=\frac{16+4\sqrt{119}}{8}
$