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

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

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

We multiply parentheses ..

(+2t^2+4t-5t-10)=0

We get rid of parentheses

2t^2+4t-5t-10=0

We add all the numbers together, and all the variables

2t^2-1t-10=0

a = 2; b = -1; c = -10;
Δ = b2-4ac
Δ = -12-4·2·(-10)
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{-(-1)-9}{2*2}=\frac{-8}{4}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*2}=\frac{10}{4}
=2+1/2
$