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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


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

-2x^2-3x+2-()

We add all the numbers together, and all the variables

-2x^2-3x

Back to the equation:

+(-2x^2-3x)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+x-1=0

a = 1; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·1·(-1)
Δ = 5
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{-(1)-\sqrt{5}}{2*1}=\frac{-1-\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5}}{2*1}=\frac{-1+\sqrt{5}}{2}
$