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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+3x-10=0

a = 1; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*1}=\frac{4}{2}
=2
$