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

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

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

We move all terms to the left:

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

We multiply parentheses

7x^2-35x-(3(x+1))=0


We calculate terms in parentheses: -(3(x+1)), so:

3(x+1)

We multiply parentheses

3x+3

Back to the equation:

-(3x+3)


We get rid of parentheses

7x^2-35x-3x-3=0

We add all the numbers together, and all the variables

7x^2-38x-3=0

a = 7; b = -38; c = -3;
Δ = b2-4ac
Δ = -382-4·7·(-3)
Δ = 1528
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:

$\sqrt{\Delta}=\sqrt{1528}=\sqrt{4*382}=\sqrt{4}*\sqrt{382}=2\sqrt{382}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{382}}{2*7}=\frac{38-2\sqrt{382}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{382}}{2*7}=\frac{38+2\sqrt{382}}{14}
$