(2t+5)(4t+1)=0

Solution for (2t+5)(4t+1)=0 equation:

(2t+5)(4t+1)=0

We multiply parentheses ..

(+8t^2+2t+20t+5)=0

We get rid of parentheses

8t^2+2t+20t+5=0

We add all the numbers together, and all the variables

8t^2+22t+5=0

a = 8; b = 22; c = +5;
Δ = b2-4ac
Δ = 222-4·8·5
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-18}{2*8}=\frac{-40}{16}
=-2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+18}{2*8}=\frac{-4}{16}
=-1/4
$