2x^2+4x=3(5+x)

Solution for 2x^2+4x=3(5+x) equation:

2x^2+4x=3(5+x)

We move all terms to the left:

2x^2+4x-(3(5+x))=0

We add all the numbers together, and all the variables

2x^2+4x-(3(x+5))=0


We calculate terms in parentheses: -(3(x+5)), so:

3(x+5)

We multiply parentheses

3x+15

Back to the equation:

-(3x+15)


We get rid of parentheses

2x^2+4x-3x-15=0

We add all the numbers together, and all the variables

2x^2+x-15=0

a = 2; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·2·(-15)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*2}=\frac{10}{4}
=2+1/2
$