0.5(t^2+4)=20t

Solution for 0.5(t^2+4)=20t equation:

0.5(t^2+4)=20t

We move all terms to the left:

0.5(t^2+4)-(20t)=0

We add all the numbers together, and all the variables

-20t+0.5(t^2+4)=0

We multiply parentheses

0.5t^2-20t+2=0

a = 0.5; b = -20; c = +2;
Δ = b2-4ac
Δ = -202-4·0.5·2
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-6\sqrt{11}}{2*0.5}=\frac{20-6\sqrt{11}}{1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+6\sqrt{11}}{2*0.5}=\frac{20+6\sqrt{11}}{1}
$