((x+5)(2x+12))/2=110

Solution for ((x+5)(2x+12))/2=110 equation:

((x+5)(2x+12))/2=110

We move all terms to the left:

((x+5)(2x+12))/2-(110)=0

We multiply parentheses ..

((+2x^2+12x+10x+60))/2-110=0

We multiply all the terms by the denominator


((+2x^2+12x+10x+60))-110*2=0


We calculate terms in parentheses: +((+2x^2+12x+10x+60)), so:

(+2x^2+12x+10x+60)

We get rid of parentheses

2x^2+12x+10x+60

We add all the numbers together, and all the variables

2x^2+22x+60

Back to the equation:

+(2x^2+22x+60)


We add all the numbers together, and all the variables

(2x^2+22x+60)-220=0

We get rid of parentheses

2x^2+22x+60-220=0

We add all the numbers together, and all the variables

2x^2+22x-160=0

a = 2; b = 22; c = -160;
Δ = b2-4ac
Δ = 222-4·2·(-160)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-42}{2*2}=\frac{-64}{4}
=-16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+42}{2*2}=\frac{20}{4}
=5
$