18x^2+6x-20=0

Solution for 18x^2+6x-20=0 equation:

18x^2+6x-20=0

a = 18; b = 6; c = -20;
Δ = b2-4ac
Δ = 62-4·18·(-20)
Δ = 1476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{41}}{2*18}=\frac{-6-6\sqrt{41}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{41}}{2*18}=\frac{-6+6\sqrt{41}}{36}
$