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

Solution for x=(2x^2+x-3)=0 equation:

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

We move all terms to the left:

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


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

(2x^2+x-3)

We get rid of parentheses

2x^2+x-3

Back to the equation:

-(2x^2+x-3)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2+3=0

a = -2; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-2)·3
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-2}=\frac{0-2\sqrt{6}}{-4}
=-\frac{2\sqrt{6}}{-4}
=-\frac{\sqrt{6}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-2}=\frac{0+2\sqrt{6}}{-4}
=\frac{2\sqrt{6}}{-4}
=\frac{\sqrt{6}}{-2}
$