2x2+(4x+6x2)=-2(3x-4)

Solution for 2x2+(4x+6x2)=-2(3x-4) equation:

2x^2+(4x+6x^2)=-2(3x-4)

We move all terms to the left:

2x^2+(4x+6x^2)-(-2(3x-4))=0

We get rid of parentheses

2x^2+6x^2+4x-(-2(3x-4))=0


We calculate terms in parentheses: -(-2(3x-4)), so:

-2(3x-4)

We multiply parentheses

-6x+8

Back to the equation:

-(-6x+8)


We add all the numbers together, and all the variables

8x^2+4x-(-6x+8)=0

We get rid of parentheses

8x^2+4x+6x-8=0

We add all the numbers together, and all the variables

8x^2+10x-8=0

a = 8; b = 10; c = -8;
Δ = b2-4ac
Δ = 102-4·8·(-8)
Δ = 356
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:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{89}}{2*8}=\frac{-10-2\sqrt{89}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{89}}{2*8}=\frac{-10+2\sqrt{89}}{16}
$