(4x^2+3x-2)=(-5x^2-4x-1)

Solution for (4x^2+3x-2)=(-5x^2-4x-1) equation:

(4x^2+3x-2)=(-5x^2-4x-1)

We move all terms to the left:

(4x^2+3x-2)-((-5x^2-4x-1))=0

We get rid of parentheses

-((-5x^2-4x-1))+4x^2+3x-2=0


We calculate terms in parentheses: -((-5x^2-4x-1)), so:

(-5x^2-4x-1)

We get rid of parentheses

-5x^2-4x-1

Back to the equation:

-(-5x^2-4x-1)


We add all the numbers together, and all the variables

4x^2-(-5x^2-4x-1)+3x-2=0

We get rid of parentheses

4x^2+5x^2+4x+3x+1-2=0

We add all the numbers together, and all the variables

9x^2+7x-1=0

a = 9; b = 7; c = -1;
Δ = b2-4ac
Δ = 72-4·9·(-1)
Δ = 85
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{85}}{2*9}=\frac{-7-\sqrt{85}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{85}}{2*9}=\frac{-7+\sqrt{85}}{18}
$