1.86t^2-15t+25=0

Solution for 1.86t^2-15t+25=0 equation:

1.86t^2-15t+25=0

a = 1.86; b = -15; c = +25;
Δ = b2-4ac
Δ = -152-4·1.86·25
Δ = 39
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{39}}{2*1.86}=\frac{15-\sqrt{39}}{3.72}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{39}}{2*1.86}=\frac{15+\sqrt{39}}{3.72}
$