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

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

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

We move all terms to the left:

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

We multiply parentheses

3x^2+42x-(x^2+14x+196)=0

We get rid of parentheses

3x^2-x^2+42x-14x-196=0

We add all the numbers together, and all the variables

2x^2+28x-196=0

a = 2; b = 28; c = -196;
Δ = b2-4ac
Δ = 282-4·2·(-196)
Δ = 2352
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:

$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28\sqrt{3}}{2*2}=\frac{-28-28\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28\sqrt{3}}{2*2}=\frac{-28+28\sqrt{3}}{4}
$