(2t-7)(t+2)=0

Solution for (2t-7)(t+2)=0 equation:

(2t-7)(t+2)=0

We multiply parentheses ..

(+2t^2+4t-7t-14)=0

We get rid of parentheses

2t^2+4t-7t-14=0

We add all the numbers together, and all the variables

2t^2-3t-14=0

a = 2; b = -3; c = -14;
Δ = b2-4ac
Δ = -32-4·2·(-14)
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{-(-3)-11}{2*2}=\frac{-8}{4}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*2}=\frac{14}{4}
=3+1/2
$