25a^2=4a^2+(9)^2

Solution for 25a^2=4a^2+(9)^2 equation:

25a^2=4a^2+(9)^2

We move all terms to the left:

25a^2-(4a^2+(9)^2)=0

We get rid of parentheses

25a^2-4a^2-9^2=0

We add all the numbers together, and all the variables

21a^2-81=0

a = 21; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·21·(-81)
Δ = 6804
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:

$\sqrt{\Delta}=\sqrt{6804}=\sqrt{324*21}=\sqrt{324}*\sqrt{21}=18\sqrt{21}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{21}}{2*21}=\frac{0-18\sqrt{21}}{42}
=-\frac{18\sqrt{21}}{42}
=-\frac{3\sqrt{21}}{7}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{21}}{2*21}=\frac{0+18\sqrt{21}}{42}
=\frac{18\sqrt{21}}{42}
=\frac{3\sqrt{21}}{7}
$