4t^2-32t-3=0

Solution for 4t^2-32t-3=0 equation:

4t^2-32t-3=0

a = 4; b = -32; c = -3;
Δ = b2-4ac
Δ = -322-4·4·(-3)
Δ = 1072
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:

$\sqrt{\Delta}=\sqrt{1072}=\sqrt{16*67}=\sqrt{16}*\sqrt{67}=4\sqrt{67}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{67}}{2*4}=\frac{32-4\sqrt{67}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{67}}{2*4}=\frac{32+4\sqrt{67}}{8}
$