3x^2-4x+10=12

Solution for 3x^2-4x+10=12 equation:

3x^2-4x+10=12

We move all terms to the left:

3x^2-4x+10-(12)=0

We add all the numbers together, and all the variables

3x^2-4x-2=0

a = 3; b = -4; c = -2;
Δ = b2-4ac
Δ = -42-4·3·(-2)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{10}}{2*3}=\frac{4-2\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*3}=\frac{4+2\sqrt{10}}{6}
$