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

Solution for (X+3)(2x-1)=7x-3x^2=0 equation:

(X+3)(2X-1)=7X-3X^2=0

We move all terms to the left:

(X+3)(2X-1)-(7X-3X^2)=0

We get rid of parentheses

3X^2-7X+(X+3)(2X-1)=0

We multiply parentheses ..

3X^2+(+2X^2-1X+6X-3)-7X=0

We get rid of parentheses

3X^2+2X^2-1X+6X-7X-3=0

We add all the numbers together, and all the variables

5X^2-2X-3=0

a = 5; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·5·(-3)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*5}=\frac{-6}{10}
=-3/5
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*5}=\frac{10}{10}
=1
$