3x^2-20=12x+93

Solution for 3x^2-20=12x+93 equation:

3x^2-20=12x+93

We move all terms to the left:

3x^2-20-(12x+93)=0

We get rid of parentheses

3x^2-12x-93-20=0

We add all the numbers together, and all the variables

3x^2-12x-113=0

a = 3; b = -12; c = -113;
Δ = b2-4ac
Δ = -122-4·3·(-113)
Δ = 1500
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:

$\sqrt{\Delta}=\sqrt{1500}=\sqrt{100*15}=\sqrt{100}*\sqrt{15}=10\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-10\sqrt{15}}{2*3}=\frac{12-10\sqrt{15}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+10\sqrt{15}}{2*3}=\frac{12+10\sqrt{15}}{6}
$