3x2-5x=(x-6)(x-7)

Solution for 3x2-5x=(x-6)(x-7) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

3x^2-((+x^2-7x-6x+42))-5x=0


We calculate terms in parentheses: -((+x^2-7x-6x+42)), so:

(+x^2-7x-6x+42)

We get rid of parentheses

x^2-7x-6x+42

We add all the numbers together, and all the variables

x^2-13x+42

Back to the equation:

-(x^2-13x+42)


We add all the numbers together, and all the variables

3x^2-5x-(x^2-13x+42)=0

We get rid of parentheses

3x^2-x^2-5x+13x-42=0

We add all the numbers together, and all the variables

2x^2+8x-42=0

a = 2; b = 8; c = -42;
Δ = b2-4ac
Δ = 82-4·2·(-42)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-20}{2*2}=\frac{-28}{4}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+20}{2*2}=\frac{12}{4}
=3
$