7+3(x+1)=2x^2+x+6

Solution for 7+3(x+1)=2x^2+x+6 equation:

7+3(x+1)=2x^2+x+6

We move all terms to the left:

7+3(x+1)-(2x^2+x+6)=0

We multiply parentheses

3x-(2x^2+x+6)+3+7=0

We get rid of parentheses

-2x^2+3x-x-6+3+7=0

We add all the numbers together, and all the variables

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

a = -2; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-2)·4
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*-2}=\frac{4}{-4}
=-1
$