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

Solution for (2t+7)(t-4)=0 equation:

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

We multiply parentheses ..

(+2t^2-8t+7t-28)=0

We get rid of parentheses

2t^2-8t+7t-28=0

We add all the numbers together, and all the variables

2t^2-1t-28=0

a = 2; b = -1; c = -28;
Δ = b2-4ac
Δ = -12-4·2·(-28)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-15}{2*2}=\frac{-14}{4}
=-3+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+15}{2*2}=\frac{16}{4}
=4
$