2x^2=(3x+7)(x+2)

Solution for 2x^2=(3x+7)(x+2) equation:

2x^2=(3x+7)(x+2)

We move all terms to the left:

2x^2-((3x+7)(x+2))=0

We multiply parentheses ..

2x^2-((+3x^2+6x+7x+14))=0


We calculate terms in parentheses: -((+3x^2+6x+7x+14)), so:

(+3x^2+6x+7x+14)

We get rid of parentheses

3x^2+6x+7x+14

We add all the numbers together, and all the variables

3x^2+13x+14

Back to the equation:

-(3x^2+13x+14)


We get rid of parentheses

2x^2-3x^2-13x-14=0

We add all the numbers together, and all the variables

-1x^2-13x-14=0

a = -1; b = -13; c = -14;
Δ = b2-4ac
Δ = -132-4·(-1)·(-14)
Δ = 113
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{113}}{2*-1}=\frac{13-\sqrt{113}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{113}}{2*-1}=\frac{13+\sqrt{113}}{-2}
$