3x^2+160x=4400

Solution for 3x^2+160x=4400 equation:

3x^2+160x=4400

We move all terms to the left:

3x^2+160x-(4400)=0

a = 3; b = 160; c = -4400;
Δ = b2-4ac
Δ = 1602-4·3·(-4400)
Δ = 78400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{78400}=280$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-280}{2*3}=\frac{-440}{6}
=-73+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+280}{2*3}=\frac{120}{6}
=20
$