(2x2+3x+4)=(x2-5x-3)

Solution for (2x2+3x+4)=(x2-5x-3) equation:

(2x^2+3x+4)=(x2-5x-3)

We move all terms to the left:

(2x^2+3x+4)-((x2-5x-3))=0

We add all the numbers together, and all the variables

-((+x^2-5x-3))+(2x^2+3x+4)=0

We get rid of parentheses

-((+x^2-5x-3))+2x^2+3x+4=0


We calculate terms in parentheses: -((+x^2-5x-3)), so:

(+x^2-5x-3)

We get rid of parentheses

x^2-5x-3

Back to the equation:

-(x^2-5x-3)


We add all the numbers together, and all the variables

2x^2+3x-(x^2-5x-3)+4=0

We get rid of parentheses

2x^2-x^2+3x+5x+3+4=0

We add all the numbers together, and all the variables

x^2+8x+7=0

a = 1; b = 8; c = +7;
Δ = b2-4ac
Δ = 82-4·1·7
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-6}{2*1}=\frac{-14}{2}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+6}{2*1}=\frac{-2}{2}
=-1
$