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

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

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

We multiply parentheses ..

(+2t^2+8t-1t-4)=0

We get rid of parentheses

2t^2+8t-1t-4=0

We add all the numbers together, and all the variables

2t^2+7t-4=0

a = 2; b = 7; c = -4;
Δ = b2-4ac
Δ = 72-4·2·(-4)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*2}=\frac{-16}{4}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*2}=\frac{2}{4}
=1/2
$