0.5t^2-10t+48=0

Solution for 0.5t^2-10t+48=0 equation:

0.5t^2-10t+48=0

a = 0.5; b = -10; c = +48;
Δ = b2-4ac
Δ = -102-4·0.5·48
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*0.5}=\frac{8}{1}
=8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*0.5}=\frac{12}{1}
=12
$