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

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

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

We multiply parentheses ..

(+16t^2-24t+4t-6)=0

We get rid of parentheses

16t^2-24t+4t-6=0

We add all the numbers together, and all the variables

16t^2-20t-6=0

a = 16; b = -20; c = -6;
Δ = b2-4ac
Δ = -202-4·16·(-6)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-28}{2*16}=\frac{-8}{32}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+28}{2*16}=\frac{48}{32}
=1+1/2
$