5t^2-32t+32=0

Solution for 5t^2-32t+32=0 equation:

5t^2-32t+32=0

a = 5; b = -32; c = +32;
Δ = b2-4ac
Δ = -322-4·5·32
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{6}}{2*5}=\frac{32-8\sqrt{6}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{6}}{2*5}=\frac{32+8\sqrt{6}}{10}
$