3x^2-4x+10=18-6x

Solution for 3x^2-4x+10=18-6x equation:

3x^2-4x+10=18-6x

We move all terms to the left:

3x^2-4x+10-(18-6x)=0

We add all the numbers together, and all the variables

3x^2-4x-(-6x+18)+10=0

We get rid of parentheses

3x^2-4x+6x-18+10=0

We add all the numbers together, and all the variables

3x^2+2x-8=0

a = 3; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·3·(-8)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*3}=\frac{8}{6}
=1+1/3
$