(2x+1)(x+3)=4

Solution for (2x+1)(x+3)=4 equation:

(2x+1)(x+3)=4

We move all terms to the left:

(2x+1)(x+3)-(4)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+7x-1=0

a = 2; b = 7; c = -1;
Δ = b2-4ac
Δ = 72-4·2·(-1)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{57}}{2*2}=\frac{-7-\sqrt{57}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{57}}{2*2}=\frac{-7+\sqrt{57}}{4}
$