X^+2x+3=(2x-1)(x+5)

Solution for X^+2x+3=(2x-1)(x+5) equation:

X^+2X+3=(2X-1)(X+5)

We move all terms to the left:

X^+2X+3-((2X-1)(X+5))=0

We add all the numbers together, and all the variables

3X-((2X-1)(X+5))+3=0

We multiply parentheses ..

-((+2X^2+10X-1X-5))+3X+3=0


We calculate terms in parentheses: -((+2X^2+10X-1X-5)), so:

(+2X^2+10X-1X-5)

We get rid of parentheses

2X^2+10X-1X-5

We add all the numbers together, and all the variables

2X^2+9X-5

Back to the equation:

-(2X^2+9X-5)


We add all the numbers together, and all the variables

3X-(2X^2+9X-5)+3=0

We get rid of parentheses

-2X^2+3X-9X+5+3=0

We add all the numbers together, and all the variables

-2X^2-6X+8=0

a = -2; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·(-2)·8
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*-2}=\frac{-4}{-4}
=1
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*-2}=\frac{16}{-4}
=-4
$