t^2+28t+194=0

Solution for t^2+28t+194=0 equation:

t^2+28t+194=0

a = 1; b = 28; c = +194;
Δ = b2-4ac
Δ = 282-4·1·194
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{2}}{2*1}=\frac{-28-2\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{2}}{2*1}=\frac{-28+2\sqrt{2}}{2}
$