5t^2-18t-35=0

Solution for 5t^2-18t-35=0 equation:

5t^2-18t-35=0

a = 5; b = -18; c = -35;
Δ = b2-4ac
Δ = -182-4·5·(-35)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-32}{2*5}=\frac{-14}{10}
=-1+2/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+32}{2*5}=\frac{50}{10}
=5
$