4a+5a^2+2a^2+a^2=0

Solution for 4a+5a^2+2a^2+a^2=0 equation:

4a+5a^2+2a^2+a^2=0

We add all the numbers together, and all the variables

8a^2+4a=0

a = 8; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·8·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*8}=\frac{-8}{16}
=-1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*8}=\frac{0}{16}
=0
$