(3x-5)(3x+2)=(x-7)(9+1)+50

Solution for (3x-5)(3x+2)=(x-7)(9+1)+50 equation:

(3x-5)(3x+2)=(x-7)(9+1)+50

We move all terms to the left:

(3x-5)(3x+2)-((x-7)(9+1)+50)=0

We add all the numbers together, and all the variables

(3x-5)(3x+2)-((x-7)10+50)=0

We multiply parentheses ..

(+9x^2+6x-15x-10)-((x-7)10+50)=0


We calculate terms in parentheses: -((x-7)10+50), so:

(x-7)10+50

We multiply parentheses

10x-70+50

We add all the numbers together, and all the variables

10x-20

Back to the equation:

-(10x-20)


We get rid of parentheses

9x^2+6x-15x-10x-10+20=0

We add all the numbers together, and all the variables

9x^2-19x+10=0

a = 9; b = -19; c = +10;
Δ = b2-4ac
Δ = -192-4·9·10
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*9}=\frac{18}{18}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*9}=\frac{20}{18}
=1+1/9
$