3x(x+14)=x^2+28x+196

Solution for 3x(x+14)=x^2+28x+196 equation:

3x(x+14)=x^2+28x+196

We move all terms to the left:

3x(x+14)-(x^2+28x+196)=0

We multiply parentheses

3x^2+42x-(x^2+28x+196)=0

We get rid of parentheses

3x^2-x^2+42x-28x-196=0

We add all the numbers together, and all the variables

2x^2+14x-196=0

a = 2; b = 14; c = -196;
Δ = b2-4ac
Δ = 142-4·2·(-196)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-42}{2*2}=\frac{-56}{4}
=-14
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+42}{2*2}=\frac{28}{4}
=7
$