(X+3)(2x+0)=40

Solution for (X+3)(2x+0)=40 equation:

(X+3)(2X+0)=40

We move all terms to the left:

(X+3)(2X+0)-(40)=0

We add all the numbers together, and all the variables

(X+3)(+2X)-40=0

We multiply parentheses ..

(+2X^2+6X)-40=0

We get rid of parentheses

2X^2+6X-40=0

a = 2; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·2·(-40)
Δ = 356
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:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{89}}{2*2}=\frac{-6-2\sqrt{89}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{89}}{2*2}=\frac{-6+2\sqrt{89}}{4}
$