x^2+4x+3x^2-5x+3=3x^2+7x+36

Solution for x^2+4x+3x^2-5x+3=3x^2+7x+36 equation:

x^2+4x+3x^2-5x+3=3x^2+7x+36

We move all terms to the left:

x^2+4x+3x^2-5x+3-(3x^2+7x+36)=0

We add all the numbers together, and all the variables

4x^2-1x-(3x^2+7x+36)+3=0

We get rid of parentheses

4x^2-3x^2-1x-7x-36+3=0

We add all the numbers together, and all the variables

x^2-8x-33=0

a = 1; b = -8; c = -33;
Δ = b2-4ac
Δ = -82-4·1·(-33)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-14}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+14}{2*1}=\frac{22}{2}
=11
$