x(3x+4)=440

Solution for x(3x+4)=440 equation:

x(3x+4)=440

We move all terms to the left:

x(3x+4)-(440)=0

We multiply parentheses

3x^2+4x-440=0

a = 3; b = 4; c = -440;
Δ = b2-4ac
Δ = 42-4·3·(-440)
Δ = 5296
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:

$\sqrt{\Delta}=\sqrt{5296}=\sqrt{16*331}=\sqrt{16}*\sqrt{331}=4\sqrt{331}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{331}}{2*3}=\frac{-4-4\sqrt{331}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{331}}{2*3}=\frac{-4+4\sqrt{331}}{6}
$