x+3x-3+2x2+8x-5=

Solution for x+3x-3+2x2+8x-5= equation:

x+3x-3+2x^2+8x-5=

We move all terms to the left:

x+3x-3+2x^2+8x-5-()=0

We add all the numbers together, and all the variables

2x^2+12x=0

a = 2; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·2·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*2}=\frac{0}{4}
=0
$