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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+5x^2+15x-2x-6)-(7(x-1))=0


We calculate terms in parentheses: -(7(x-1)), so:

7(x-1)

We multiply parentheses

7x-7

Back to the equation:

-(7x-7)


We get rid of parentheses

5x^2+15x-2x-7x-6+7=0

We add all the numbers together, and all the variables

5x^2+6x+1=0

a = 5; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·5·1
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*5}=\frac{-10}{10}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*5}=\frac{-2}{10}
=-1/5
$