50t^2+15t+1=0

Solution for 50t^2+15t+1=0 equation:

50t^2+15t+1=0

a = 50; b = 15; c = +1;
Δ = b2-4ac
Δ = 152-4·50·1
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5}{2*50}=\frac{-20}{100}
=-1/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5}{2*50}=\frac{-10}{100}
=-1/10
$