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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((3x^2+4x-5)), so:

(3x^2+4x-5)

We get rid of parentheses

3x^2+4x-5

Back to the equation:

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


We get rid of parentheses

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

We add all the numbers together, and all the variables

-5x^2-4x+8=0

a = -5; b = -4; c = +8;
Δ = b2-4ac
Δ = -42-4·(-5)·8
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{11}}{2*-5}=\frac{4-4\sqrt{11}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{11}}{2*-5}=\frac{4+4\sqrt{11}}{-10}
$