(2t-7)(5t+6)=0

Solution for (2t-7)(5t+6)=0 equation:

(2t-7)(5t+6)=0

We multiply parentheses ..

(+10t^2+12t-35t-42)=0

We get rid of parentheses

10t^2+12t-35t-42=0

We add all the numbers together, and all the variables

10t^2-23t-42=0

a = 10; b = -23; c = -42;
Δ = b2-4ac
Δ = -232-4·10·(-42)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-47}{2*10}=\frac{-24}{20}
=-1+1/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+47}{2*10}=\frac{70}{20}
=3+1/2
$