(2t-10t2)+14(4t+2)=0

Solution for (2t-10t2)+14(4t+2)=0 equation:

(2t-10t^2)+14(4t+2)=0

We multiply parentheses

(2t-10t^2)+56t+28=0

We get rid of parentheses

-10t^2+2t+56t+28=0

We add all the numbers together, and all the variables

-10t^2+58t+28=0

a = -10; b = 58; c = +28;
Δ = b2-4ac
Δ = 582-4·(-10)·28
Δ = 4484
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:

$\sqrt{\Delta}=\sqrt{4484}=\sqrt{4*1121}=\sqrt{4}*\sqrt{1121}=2\sqrt{1121}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-2\sqrt{1121}}{2*-10}=\frac{-58-2\sqrt{1121}}{-20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+2\sqrt{1121}}{2*-10}=\frac{-58+2\sqrt{1121}}{-20}
$