4t^2-16t-52=0

Solution for 4t^2-16t-52=0 equation:

4t^2-16t-52=0

a = 4; b = -16; c = -52;
Δ = b2-4ac
Δ = -162-4·4·(-52)
Δ = 1088
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:

$\sqrt{\Delta}=\sqrt{1088}=\sqrt{64*17}=\sqrt{64}*\sqrt{17}=8\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{17}}{2*4}=\frac{16-8\sqrt{17}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{17}}{2*4}=\frac{16+8\sqrt{17}}{8}
$