2x^2+4x+6x=13

Solution for 2x^2+4x+6x=13 equation:

2x^2+4x+6x=13

We move all terms to the left:

2x^2+4x+6x-(13)=0

We add all the numbers together, and all the variables

2x^2+10x-13=0

a = 2; b = 10; c = -13;
Δ = b2-4ac
Δ = 102-4·2·(-13)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{51}}{2*2}=\frac{-10-2\sqrt{51}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{51}}{2*2}=\frac{-10+2\sqrt{51}}{4}
$