5t^2-29t-42=0

Solution for 5t^2-29t-42=0 equation:

5t^2-29t-42=0

a = 5; b = -29; c = -42;
Δ = b2-4ac
Δ = -292-4·5·(-42)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-41}{2*5}=\frac{-12}{10}
=-1+1/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+41}{2*5}=\frac{70}{10}
=7
$