(2x2+x-4)=(-3x^2-5x+1)

Solution for (2x2+x-4)=(-3x^2-5x+1) equation:

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((-3x^2-5x+1)), so:

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

We get rid of parentheses

-3x^2-5x+1

Back to the equation:

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


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5x^2+6x-5=0

a = 5; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·5·(-5)
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{34}}{2*5}=\frac{-6-2\sqrt{34}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{34}}{2*5}=\frac{-6+2\sqrt{34}}{10}
$