1a^2=176+27a

Solution for 1a^2=176+27a equation:

1a^2=176+27a

We move all terms to the left:

1a^2-(176+27a)=0

We add all the numbers together, and all the variables

1a^2-(27a+176)=0

We add all the numbers together, and all the variables

a^2-(27a+176)=0

We get rid of parentheses

a^2-27a-176=0

a = 1; b = -27; c = -176;
Δ = b2-4ac
Δ = -272-4·1·(-176)
Δ = 1433
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-\sqrt{1433}}{2*1}=\frac{27-\sqrt{1433}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{1433}}{2*1}=\frac{27+\sqrt{1433}}{2}
$