18x^2+51x=41

Solution for 18x^2+51x=41 equation:

18x^2+51x=41

We move all terms to the left:

18x^2+51x-(41)=0

a = 18; b = 51; c = -41;
Δ = b2-4ac
Δ = 512-4·18·(-41)
Δ = 5553
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:

$\sqrt{\Delta}=\sqrt{5553}=\sqrt{9*617}=\sqrt{9}*\sqrt{617}=3\sqrt{617}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-3\sqrt{617}}{2*18}=\frac{-51-3\sqrt{617}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+3\sqrt{617}}{2*18}=\frac{-51+3\sqrt{617}}{36}
$