0.3t^2+60t+75=0

Solution for 0.3t^2+60t+75=0 equation:

0.3t^2+60t+75=0

a = 0.3; b = 60; c = +75;
Δ = b2-4ac
Δ = 602-4·0.3·75
Δ = 3510
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:

$\sqrt{\Delta}=\sqrt{3510}=\sqrt{9*390}=\sqrt{9}*\sqrt{390}=3\sqrt{390}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-3\sqrt{390}}{2*0.3}=\frac{-60-3\sqrt{390}}{0.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+3\sqrt{390}}{2*0.3}=\frac{-60+3\sqrt{390}}{0.6}
$