(2t-5)(t+3)=0

Solution for (2t-5)(t+3)=0 equation:

(2t-5)(t+3)=0

We multiply parentheses ..

(+2t^2+6t-5t-15)=0

We get rid of parentheses

2t^2+6t-5t-15=0

We add all the numbers together, and all the variables

2t^2+t-15=0

a = 2; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·2·(-15)
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{-(1)-11}{2*2}=\frac{-12}{4}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*2}=\frac{10}{4}
=2+1/2
$