49t^2-160t+50=0

Solution for 49t^2-160t+50=0 equation:

49t^2-160t+50=0

a = 49; b = -160; c = +50;
Δ = b2-4ac
Δ = -1602-4·49·50
Δ = 15800
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:

$\sqrt{\Delta}=\sqrt{15800}=\sqrt{100*158}=\sqrt{100}*\sqrt{158}=10\sqrt{158}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-10\sqrt{158}}{2*49}=\frac{160-10\sqrt{158}}{98}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+10\sqrt{158}}{2*49}=\frac{160+10\sqrt{158}}{98}
$