(2t-1)(5t-1)-13=0

Solution for (2t-1)(5t-1)-13=0 equation:

(2t-1)(5t-1)-13=0

We multiply parentheses ..

(+10t^2-2t-5t+1)-13=0

We get rid of parentheses

10t^2-2t-5t+1-13=0

We add all the numbers together, and all the variables

10t^2-7t-12=0

a = 10; b = -7; c = -12;
Δ = b2-4ac
Δ = -72-4·10·(-12)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-23}{2*10}=\frac{-16}{20}
=-4/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+23}{2*10}=\frac{30}{20}
=1+1/2
$