10+((a^2)/9)=16

Solution for 10+((a^2)/9)=16 equation:

10+((a^2)/9)=16

We move all terms to the left:

10+((a^2)/9)-(16)=0

We add all the numbers together, and all the variables

(a^2/9)-6=0

We get rid of parentheses

a^2/9-6=0

We multiply all the terms by the denominator


a^2-6*9=0

We add all the numbers together, and all the variables

a^2-54=0

a = 1; b = 0; c = -54;
Δ = b2-4ac
Δ = 02-4·1·(-54)
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*1}=\frac{0-6\sqrt{6}}{2}
=-\frac{6\sqrt{6}}{2}
=-3\sqrt{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*1}=\frac{0+6\sqrt{6}}{2}
=\frac{6\sqrt{6}}{2}
=3\sqrt{6}
$