-2(2x-3)=3x(x+1)+10

Solution for -2(2x-3)=3x(x+1)+10 equation:

-2(2x-3)=3x(x+1)+10

We move all terms to the left:

-2(2x-3)-(3x(x+1)+10)=0

We multiply parentheses

-4x-(3x(x+1)+10)+6=0


We calculate terms in parentheses: -(3x(x+1)+10), so:

3x(x+1)+10

We multiply parentheses

3x^2+3x+10

Back to the equation:

-(3x^2+3x+10)


We get rid of parentheses

-3x^2-4x-3x-10+6=0

We add all the numbers together, and all the variables

-3x^2-7x-4=0

a = -3; b = -7; c = -4;
Δ = b2-4ac
Δ = -72-4·(-3)·(-4)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*-3}=\frac{6}{-6}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*-3}=\frac{8}{-6}
=-1+1/3
$