1.75t^2+30t-100=0

Solution for 1.75t^2+30t-100=0 equation:

1.75t^2+30t-100=0

a = 1.75; b = 30; c = -100;
Δ = b2-4ac
Δ = 302-4·1.75·(-100)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-40}{2*1.75}=\frac{-70}{3.5}
=-20
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+40}{2*1.75}=\frac{10}{3.5}
=2+1/1.16666666667
$