-(8t+3)(2t-5)=0

Solution for -(8t+3)(2t-5)=0 equation:

-(8t+3)(2t-5)=0

We multiply parentheses ..

-(+16t^2-40t+6t-15)=0

We get rid of parentheses

-16t^2+40t-6t+15=0

We add all the numbers together, and all the variables

-16t^2+34t+15=0

a = -16; b = 34; c = +15;
Δ = b2-4ac
Δ = 342-4·(-16)·15
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-46}{2*-16}=\frac{-80}{-32}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+46}{2*-16}=\frac{12}{-32}
=-3/8
$