37x^2+12x+1=18

Solution for 37x^2+12x+1=18 equation:

37x^2+12x+1=18

We move all terms to the left:

37x^2+12x+1-(18)=0

We add all the numbers together, and all the variables

37x^2+12x-17=0

a = 37; b = 12; c = -17;
Δ = b2-4ac
Δ = 122-4·37·(-17)
Δ = 2660
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:

$\sqrt{\Delta}=\sqrt{2660}=\sqrt{4*665}=\sqrt{4}*\sqrt{665}=2\sqrt{665}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{665}}{2*37}=\frac{-12-2\sqrt{665}}{74}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{665}}{2*37}=\frac{-12+2\sqrt{665}}{74}
$