15t2+4t-35=0

Solution for 15t2+4t-35=0 equation:

15t^2+4t-35=0

a = 15; b = 4; c = -35;
Δ = b2-4ac
Δ = 42-4·15·(-35)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-46}{2*15}=\frac{-50}{30}
=-1+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+46}{2*15}=\frac{42}{30}
=1+2/5
$