(8t-5)(2t+6)=0

Solution for (8t-5)(2t+6)=0 equation:

(8t-5)(2t+6)=0

We multiply parentheses ..

(+16t^2+48t-10t-30)=0

We get rid of parentheses

16t^2+48t-10t-30=0

We add all the numbers together, and all the variables

16t^2+38t-30=0

a = 16; b = 38; c = -30;
Δ = b2-4ac
Δ = 382-4·16·(-30)
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-58}{2*16}=\frac{-96}{32}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+58}{2*16}=\frac{20}{32}
=5/8
$