3x^2=(x+10)(x+12)

Solution for 3x^2=(x+10)(x+12) equation:

3x^2=(x+10)(x+12)

We move all terms to the left:

3x^2-((x+10)(x+12))=0

We multiply parentheses ..

3x^2-((+x^2+12x+10x+120))=0


We calculate terms in parentheses: -((+x^2+12x+10x+120)), so:

(+x^2+12x+10x+120)

We get rid of parentheses

x^2+12x+10x+120

We add all the numbers together, and all the variables

x^2+22x+120

Back to the equation:

-(x^2+22x+120)


We get rid of parentheses

3x^2-x^2-22x-120=0

We add all the numbers together, and all the variables

2x^2-22x-120=0

a = 2; b = -22; c = -120;
Δ = b2-4ac
Δ = -222-4·2·(-120)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-38}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+38}{2*2}=\frac{60}{4}
=15
$