18x^2+10x-3=0

Solution for 18x^2+10x-3=0 equation:

18x^2+10x-3=0

a = 18; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·18·(-3)
Δ = 316
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:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{79}}{2*18}=\frac{-10-2\sqrt{79}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{79}}{2*18}=\frac{-10+2\sqrt{79}}{36}
$