18x^2+35x+17=0

Solution for 18x^2+35x+17=0 equation:

18x^2+35x+17=0

a = 18; b = 35; c = +17;
Δ = b2-4ac
Δ = 352-4·18·17
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-1}{2*18}=\frac{-36}{36}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+1}{2*18}=\frac{-34}{36}
=-17/18
$