16t^2-198t+46=0

Solution for 16t^2-198t+46=0 equation:

16t^2-198t+46=0

a = 16; b = -198; c = +46;
Δ = b2-4ac
Δ = -1982-4·16·46
Δ = 36260
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:

$\sqrt{\Delta}=\sqrt{36260}=\sqrt{196*185}=\sqrt{196}*\sqrt{185}=14\sqrt{185}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-198)-14\sqrt{185}}{2*16}=\frac{198-14\sqrt{185}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-198)+14\sqrt{185}}{2*16}=\frac{198+14\sqrt{185}}{32}
$