4t^2-24t+11=0

Solution for 4t^2-24t+11=0 equation:

4t^2-24t+11=0

a = 4; b = -24; c = +11;
Δ = b2-4ac
Δ = -242-4·4·11
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-20}{2*4}=\frac{4}{8}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+20}{2*4}=\frac{44}{8}
=5+1/2
$