3x^2+92x=-9

Solution for 3x^2+92x=-9 equation:

3x^2+92x=-9

We move all terms to the left:

3x^2+92x-(-9)=0

We add all the numbers together, and all the variables

3x^2+92x+9=0

a = 3; b = 92; c = +9;
Δ = b2-4ac
Δ = 922-4·3·9
Δ = 8356
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:

$\sqrt{\Delta}=\sqrt{8356}=\sqrt{4*2089}=\sqrt{4}*\sqrt{2089}=2\sqrt{2089}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(92)-2\sqrt{2089}}{2*3}=\frac{-92-2\sqrt{2089}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(92)+2\sqrt{2089}}{2*3}=\frac{-92+2\sqrt{2089}}{6}
$