6(2x-1)-12=-2(x2+10)

Solution for 6(2x-1)-12=-2(x2+10) equation:

6(2x-1)-12=-2(x2+10)

We move all terms to the left:

6(2x-1)-12-(-2(x2+10))=0

We add all the numbers together, and all the variables

-(-2(+x^2+10))+6(2x-1)-12=0

We multiply parentheses

-(-2(+x^2+10))+12x-6-12=0


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

-2(+x^2+10)

We multiply parentheses

-2x^2-20

Back to the equation:

-(-2x^2-20)


We add all the numbers together, and all the variables

-(-2x^2-20)+12x-18=0

We get rid of parentheses

2x^2+12x+20-18=0

We add all the numbers together, and all the variables

2x^2+12x+2=0

a = 2; b = 12; c = +2;
Δ = b2-4ac
Δ = 122-4·2·2
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{2}}{2*2}=\frac{-12-8\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{2}}{2*2}=\frac{-12+8\sqrt{2}}{4}
$