117x^2+40x+3=180

Solution for 117x^2+40x+3=180 equation:

117x^2+40x+3=180

We move all terms to the left:

117x^2+40x+3-(180)=0

We add all the numbers together, and all the variables

117x^2+40x-177=0

a = 117; b = 40; c = -177;
Δ = b2-4ac
Δ = 402-4·117·(-177)
Δ = 84436
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:

$\sqrt{\Delta}=\sqrt{84436}=\sqrt{4*21109}=\sqrt{4}*\sqrt{21109}=2\sqrt{21109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{21109}}{2*117}=\frac{-40-2\sqrt{21109}}{234}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{21109}}{2*117}=\frac{-40+2\sqrt{21109}}{234}
$