(8t+6)(2t-1)=0

Solution for (8t+6)(2t-1)=0 equation:

(8t+6)(2t-1)=0

We multiply parentheses ..

(+16t^2-8t+12t-6)=0

We get rid of parentheses

16t^2-8t+12t-6=0

We add all the numbers together, and all the variables

16t^2+4t-6=0

a = 16; b = 4; c = -6;
Δ = b2-4ac
Δ = 42-4·16·(-6)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*16}=\frac{-24}{32}
=-3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*16}=\frac{16}{32}
=1/2
$