(4x-9+)(3x+7)=(6x+15)

Solution for (4x-9+)(3x+7)=(6x+15) equation:

(4x-9+)(3x+7)=(6x+15)

We move all terms to the left:

(4x-9+)(3x+7)-((6x+15))=0

We add all the numbers together, and all the variables

(+4x)(3x+7)-((6x+15))=0

We multiply parentheses ..

(+12x^2+28x)-((6x+15))=0


We calculate terms in parentheses: -((6x+15)), so:

(6x+15)

We get rid of parentheses

6x+15

Back to the equation:

-(6x+15)


We get rid of parentheses

12x^2+28x-6x-15=0

We add all the numbers together, and all the variables

12x^2+22x-15=0

a = 12; b = 22; c = -15;
Δ = b2-4ac
Δ = 222-4·12·(-15)
Δ = 1204
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1204}=\sqrt{4*301}=\sqrt{4}*\sqrt{301}=2\sqrt{301}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{301}}{2*12}=\frac{-22-2\sqrt{301}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{301}}{2*12}=\frac{-22+2\sqrt{301}}{24}
$