5t^2+20t=300

Solution for 5t^2+20t=300 equation:

5t^2+20t=300

We move all terms to the left:

5t^2+20t-(300)=0

a = 5; b = 20; c = -300;
Δ = b2-4ac
Δ = 202-4·5·(-300)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-80}{2*5}=\frac{-100}{10}
=-10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+80}{2*5}=\frac{60}{10}
=6
$