3x^2-8x+35=2x^2+4x

Solution for 3x^2-8x+35=2x^2+4x equation:

3x^2-8x+35=2x^2+4x

We move all terms to the left:

3x^2-8x+35-(2x^2+4x)=0

We get rid of parentheses

3x^2-2x^2-8x-4x+35=0

We add all the numbers together, and all the variables

x^2-12x+35=0

a = 1; b = -12; c = +35;
Δ = b2-4ac
Δ = -122-4·1·35
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2}{2*1}=\frac{10}{2}
=5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2}{2*1}=\frac{14}{2}
=7
$