x-(2x-3)=-6(x^2+x-2)

Solution for x-(2x-3)=-6(x^2+x-2) equation:

x-(2x-3)=-6(x^2+x-2)

We move all terms to the left:

x-(2x-3)-(-6(x^2+x-2))=0

We get rid of parentheses

x-2x-(-6(x^2+x-2))+3=0


We calculate terms in parentheses: -(-6(x^2+x-2)), so:

-6(x^2+x-2)

We multiply parentheses

-6x^2-6x+12

Back to the equation:

-(-6x^2-6x+12)


We add all the numbers together, and all the variables

-(-6x^2-6x+12)-1x+3=0

We get rid of parentheses

6x^2+6x-1x-12+3=0

We add all the numbers together, and all the variables

6x^2+5x-9=0

a = 6; b = 5; c = -9;
Δ = b2-4ac
Δ = 52-4·6·(-9)
Δ = 241
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{-(5)-\sqrt{241}}{2*6}=\frac{-5-\sqrt{241}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{241}}{2*6}=\frac{-5+\sqrt{241}}{12}
$