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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3x^2-10x-(-(x+5)(-1x+2))-4=0

We multiply parentheses ..

3x^2-(-(-1x^2+2x-5x+10))-10x-4=0


We calculate terms in parentheses: -(-(-1x^2+2x-5x+10)), so:

-(-1x^2+2x-5x+10)

We get rid of parentheses

1x^2-2x+5x-10

We add all the numbers together, and all the variables

x^2+3x-10

Back to the equation:

-(x^2+3x-10)


We add all the numbers together, and all the variables

3x^2-10x-(x^2+3x-10)-4=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-13x+6=0

a = 2; b = -13; c = +6;
Δ = b2-4ac
Δ = -132-4·2·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*2}=\frac{2}{4}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*2}=\frac{24}{4}
=6
$