3(x^2+3)=28x

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

3(x^2+3)=28x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3x^2-28x+9=0

a = 3; b = -28; c = +9;
Δ = b2-4ac
Δ = -282-4·3·9
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-26}{2*3}=\frac{2}{6}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+26}{2*3}=\frac{54}{6}
=9
$