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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+42x-224=0

a = 2; b = 42; c = -224;
Δ = b2-4ac
Δ = 422-4·2·(-224)
Δ = 3556
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:

$\sqrt{\Delta}=\sqrt{3556}=\sqrt{4*889}=\sqrt{4}*\sqrt{889}=2\sqrt{889}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{889}}{2*2}=\frac{-42-2\sqrt{889}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{889}}{2*2}=\frac{-42+2\sqrt{889}}{4}
$