-5t^2+20t+5=10

Solution for -5t^2+20t+5=10 equation:

-5t^2+20t+5=10

We move all terms to the left:

-5t^2+20t+5-(10)=0

We add all the numbers together, and all the variables

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

a = -5; b = 20; c = -5;
Δ = b2-4ac
Δ = 202-4·(-5)·(-5)
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{3}}{2*-5}=\frac{-20-10\sqrt{3}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{3}}{2*-5}=\frac{-20+10\sqrt{3}}{-10}
$