10=-4t+2t^2

Solution for 10=-4t+2t^2 equation:

10=-4t+2t^2

We move all terms to the left:

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

We get rid of parentheses

-2t^2+4t+10=0

a = -2; b = 4; c = +10;
Δ = b2-4ac
Δ = 42-4·(-2)·10
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{6}}{2*-2}=\frac{-4-4\sqrt{6}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{6}}{2*-2}=\frac{-4+4\sqrt{6}}{-4}
$