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

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

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

We multiply parentheses ..

(+2t^2+10t-1t-5)=0

We get rid of parentheses

2t^2+10t-1t-5=0

We add all the numbers together, and all the variables

2t^2+9t-5=0

a = 2; b = 9; c = -5;
Δ = b2-4ac
Δ = 92-4·2·(-5)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-11}{2*2}=\frac{-20}{4}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+11}{2*2}=\frac{2}{4}
=1/2
$