15t^2-32t+11=0

Solution for 15t^2-32t+11=0 equation:

15t^2-32t+11=0

a = 15; b = -32; c = +11;
Δ = b2-4ac
Δ = -322-4·15·11
Δ = 364
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:

$\sqrt{\Delta}=\sqrt{364}=\sqrt{4*91}=\sqrt{4}*\sqrt{91}=2\sqrt{91}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{91}}{2*15}=\frac{32-2\sqrt{91}}{30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{91}}{2*15}=\frac{32+2\sqrt{91}}{30}
$