7(x^2-6x+9)=5(x-3)

Solution for 7(x^2-6x+9)=5(x-3) equation:

7(x^2-6x+9)=5(x-3)

We move all terms to the left:

7(x^2-6x+9)-(5(x-3))=0

We multiply parentheses

7x^2-42x-(5(x-3))+63=0


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

5(x-3)

We multiply parentheses

5x-15

Back to the equation:

-(5x-15)


We get rid of parentheses

7x^2-42x-5x+15+63=0

We add all the numbers together, and all the variables

7x^2-47x+78=0

a = 7; b = -47; c = +78;
Δ = b2-4ac
Δ = -472-4·7·78
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{-(-47)-5}{2*7}=\frac{42}{14}
=3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+5}{2*7}=\frac{52}{14}
=3+5/7
$