25t^2-70t+48=0

Solution for 25t^2-70t+48=0 equation:

25t^2-70t+48=0

a = 25; b = -70; c = +48;
Δ = b2-4ac
Δ = -702-4·25·48
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-70)-10}{2*25}=\frac{60}{50}
=1+1/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+10}{2*25}=\frac{80}{50}
=1+3/5
$