3x^2+5x+2x^2+3x=+

Solution for 3x^2+5x+2x^2+3x=+ equation:

3x^2+5x+2x^2+3x=+

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

5x^2+8x=0

a = 5; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·5·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*5}=\frac{-16}{10}
=-1+3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*5}=\frac{0}{10}
=0
$