3x^2+16x=204

Solution for 3x^2+16x=204 equation:

3x^2+16x=204

We move all terms to the left:

3x^2+16x-(204)=0

a = 3; b = 16; c = -204;
Δ = b2-4ac
Δ = 162-4·3·(-204)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-52}{2*3}=\frac{-68}{6}
=-11+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+52}{2*3}=\frac{36}{6}
=6
$