(2t)^2-5(2t)-14=0

Solution for (2t)^2-5(2t)-14=0 equation:

(2t)^2-5(2t)-14=0

a = 2; b = -52; c = -14;
Δ = b2-4ac
Δ = -522-4·2·(-14)
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-16\sqrt{11}}{2*2}=\frac{52-16\sqrt{11}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+16\sqrt{11}}{2*2}=\frac{52+16\sqrt{11}}{4}
$