15t^2-26t+7=0

Solution for 15t^2-26t+7=0 equation:

15t^2-26t+7=0

a = 15; b = -26; c = +7;
Δ = b2-4ac
Δ = -262-4·15·7
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-16}{2*15}=\frac{10}{30}
=1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+16}{2*15}=\frac{42}{30}
=1+2/5
$