18x+48=3x^2+6x

Solution for 18x+48=3x^2+6x equation:

18x+48=3x^2+6x

We move all terms to the left:

18x+48-(3x^2+6x)=0

We get rid of parentheses

-3x^2+18x-6x+48=0

We add all the numbers together, and all the variables

-3x^2+12x+48=0

a = -3; b = 12; c = +48;
Δ = b2-4ac
Δ = 122-4·(-3)·48
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{5}}{2*-3}=\frac{-12-12\sqrt{5}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{5}}{2*-3}=\frac{-12+12\sqrt{5}}{-6}
$