2x2-(x-2)(x+5)=7(x+3)

Solution for 2x2-(x-2)(x+5)=7(x+3) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

2x^2-(+x^2+5x-2x-10)-(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

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

We add all the numbers together, and all the variables

x^2-10x-11=0

a = 1; b = -10; c = -11;
Δ = b2-4ac
Δ = -102-4·1·(-11)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-12}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+12}{2*1}=\frac{22}{2}
=11
$