t=5t-10t^2+10

Solution for t=5t-10t^2+10 equation:

t=5t-10t^2+10

We move all terms to the left:

t-(5t-10t^2+10)=0

We get rid of parentheses

10t^2-5t+t-10=0

We add all the numbers together, and all the variables

10t^2-4t-10=0

a = 10; b = -4; c = -10;
Δ = b2-4ac
Δ = -42-4·10·(-10)
Δ = 416
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:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{26}}{2*10}=\frac{4-4\sqrt{26}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{26}}{2*10}=\frac{4+4\sqrt{26}}{20}
$