-5t^2+5t+20=t

Solution for -5t^2+5t+20=t equation:

-5t^2+5t+20=t

We move all terms to the left:

-5t^2+5t+20-(t)=0

We add all the numbers together, and all the variables

-5t^2+4t+20=0

a = -5; b = 4; c = +20;
Δ = b2-4ac
Δ = 42-4·(-5)·20
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*-5}=\frac{-4-4\sqrt{26}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{26}}{2*-5}=\frac{-4+4\sqrt{26}}{-10}
$