3x^2+x=2x+10

Solution for 3x^2+x=2x+10 equation:

3x^2+x=2x+10

We move all terms to the left:

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

We get rid of parentheses

3x^2+x-2x-10=0

We add all the numbers together, and all the variables

3x^2-1x-10=0

a = 3; b = -1; c = -10;
Δ = b2-4ac
Δ = -12-4·3·(-10)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*3}=\frac{-10}{6}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*3}=\frac{12}{6}
=2
$