10t^2+17t-20=0

Solution for 10t^2+17t-20=0 equation:

10t^2+17t-20=0

a = 10; b = 17; c = -20;
Δ = b2-4ac
Δ = 172-4·10·(-20)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-33}{2*10}=\frac{-50}{20}
=-2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+33}{2*10}=\frac{16}{20}
=4/5
$