A=2x^2+10x+8

Solution for A=2x^2+10x+8 equation:

=2A^2+10A+8

We move all terms to the left:

-(2A^2+10A+8)=0

We get rid of parentheses

-2A^2-10A-8=0

a = -2; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·(-2)·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6}{2*-2}=\frac{4}{-4}
=-1
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6}{2*-2}=\frac{16}{-4}
=-4
$