(x+4)(2x+12)=16(x+18)

Solution for (x+4)(2x+12)=16(x+18) equation:

(x+4)(2x+12)=16(x+18)

We move all terms to the left:

(x+4)(2x+12)-(16(x+18))=0

We multiply parentheses ..

(+2x^2+12x+8x+48)-(16(x+18))=0


We calculate terms in parentheses: -(16(x+18)), so:

16(x+18)

We multiply parentheses

16x+288

Back to the equation:

-(16x+288)


We get rid of parentheses

2x^2+12x+8x-16x+48-288=0

We add all the numbers together, and all the variables

2x^2+4x-240=0

a = 2; b = 4; c = -240;
Δ = b2-4ac
Δ = 42-4·2·(-240)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-44}{2*2}=\frac{-48}{4}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+44}{2*2}=\frac{40}{4}
=10
$