t^2+16t+11=0

Solution for t^2+16t+11=0 equation:

t^2+16t+11=0

a = 1; b = 16; c = +11;
Δ = b2-4ac
Δ = 162-4·1·11
Δ = 212
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:

$\sqrt{\Delta}=\sqrt{212}=\sqrt{4*53}=\sqrt{4}*\sqrt{53}=2\sqrt{53}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{53}}{2*1}=\frac{-16-2\sqrt{53}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{53}}{2*1}=\frac{-16+2\sqrt{53}}{2}
$