7x^2+2(2x+3)=2(3x^2-4)+13x

Solution for 7x^2+2(2x+3)=2(3x^2-4)+13x equation:

7x^2+2(2x+3)=2(3x^2-4)+13x

We move all terms to the left:

7x^2+2(2x+3)-(2(3x^2-4)+13x)=0

We multiply parentheses

7x^2+4x-(2(3x^2-4)+13x)+6=0


We calculate terms in parentheses: -(2(3x^2-4)+13x), so:

2(3x^2-4)+13x

We add all the numbers together, and all the variables

13x+2(3x^2-4)

We multiply parentheses

6x^2+13x-8

Back to the equation:

-(6x^2+13x-8)


We get rid of parentheses

7x^2-6x^2+4x-13x+8+6=0

We add all the numbers together, and all the variables

x^2-9x+14=0

a = 1; b = -9; c = +14;
Δ = b2-4ac
Δ = -92-4·1·14
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-5}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+5}{2*1}=\frac{14}{2}
=7
$