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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -((x+3)x), so:

(x+3)x

We multiply parentheses

x^2+3x

Back to the equation:

-(x^2+3x)


We get rid of parentheses

2x^2-x^2-3x=0

We add all the numbers together, and all the variables

x^2-3x=0

a = 1; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·1·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*1}=\frac{6}{2}
=3
$