(x+3)(2x-4)=5

Solution for (x+3)(2x-4)=5 equation:

(x+3)(2x-4)=5

We move all terms to the left:

(x+3)(2x-4)-(5)=0

We multiply parentheses ..

(+2x^2-4x+6x-12)-5=0

We get rid of parentheses

2x^2-4x+6x-12-5=0

We add all the numbers together, and all the variables

2x^2+2x-17=0

a = 2; b = 2; c = -17;
Δ = b2-4ac
Δ = 22-4·2·(-17)
Δ = 140
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{35}}{2*2}=\frac{-2-2\sqrt{35}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{35}}{2*2}=\frac{-2+2\sqrt{35}}{4}
$