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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-(+x^2+5x-2x-10)+2x-(7(x+3))=0


We calculate terms in parentheses: -(7(x+3)), so:

7(x+3)

We multiply parentheses

7x+21

Back to the equation:

-(7x+21)


We get rid of parentheses

-x^2-5x+2x+2x-7x+10-21=0

We add all the numbers together, and all the variables

-1x^2-8x-11=0

a = -1; b = -8; c = -11;
Δ = b2-4ac
Δ = -82-4·(-1)·(-11)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{5}}{2*-1}=\frac{8-2\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{5}}{2*-1}=\frac{8+2\sqrt{5}}{-2}
$