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

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

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

We move all terms to the left:

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

We multiply parentheses ..

2(+x^2-4x+3x-12)-(7(x+3))=0


We calculate terms in parentheses: -(7(x+3)), so:

7(x+3)

We multiply parentheses

7x+21

Back to the equation:

-(7x+21)


We multiply parentheses

2x^2-8x+6x-(7x+21)-24=0

We get rid of parentheses

2x^2-8x+6x-7x-21-24=0

We add all the numbers together, and all the variables

2x^2-9x-45=0

a = 2; b = -9; c = -45;
Δ = b2-4ac
Δ = -92-4·2·(-45)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-21}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+21}{2*2}=\frac{30}{4}
=7+1/2
$