t^2-16t-28=0

Solution for t^2-16t-28=0 equation:

t^2-16t-28=0

a = 1; b = -16; c = -28;
Δ = b2-4ac
Δ = -162-4·1·(-28)
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{23}}{2*1}=\frac{16-4\sqrt{23}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{23}}{2*1}=\frac{16+4\sqrt{23}}{2}
$