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

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

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

We move all terms to the left:

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


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

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

We get rid of parentheses

-2x^2+5x+3

Back to the equation:

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



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

(3x^2+5x-3)

We get rid of parentheses

3x^2+5x-3

Back to the equation:

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


We get rid of parentheses

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

We add all the numbers together, and all the variables

-5x^2+6=0

a = -5; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-5)·6
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*-5}=\frac{0-2\sqrt{30}}{-10}
=-\frac{2\sqrt{30}}{-10}
=-\frac{\sqrt{30}}{-5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*-5}=\frac{0+2\sqrt{30}}{-10}
=\frac{2\sqrt{30}}{-10}
=\frac{\sqrt{30}}{-5}
$