7(x-3)+(x-3)(5x+4)=0

Solution for 7(x-3)+(x-3)(5x+4)=0 equation:

7(x-3)+(x-3)(5x+4)=0

We multiply parentheses

7x+(x-3)(5x+4)-21=0

We multiply parentheses ..

(+5x^2+4x-15x-12)+7x-21=0

We get rid of parentheses

5x^2+4x-15x+7x-12-21=0

We add all the numbers together, and all the variables

5x^2-4x-33=0

a = 5; b = -4; c = -33;
Δ = b2-4ac
Δ = -42-4·5·(-33)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-26}{2*5}=\frac{-22}{10}
=-2+1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+26}{2*5}=\frac{30}{10}
=3
$