7^(2x)x5^(3x)=21/25

Solution for 7^(2x)x5^(3x)=21/25 equation:

7^(2x)x5^(3x)=21/25

We move all terms to the left:

7^(2x)x5^(3x)-(21/25)=0

We add all the numbers together, and all the variables

7^2xx5^3x-(+21/25)=0

We get rid of parentheses

7^2xx5^3x-21/25=0

We multiply all the terms by the denominator


7^2xx5^3x*25-21=0

Wy multiply elements

175x^2-21=0

a = 175; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·175·(-21)
Δ = 14700
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{14700}=\sqrt{4900*3}=\sqrt{4900}*\sqrt{3}=70\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-70\sqrt{3}}{2*175}=\frac{0-70\sqrt{3}}{350}
=-\frac{70\sqrt{3}}{350}
=-\frac{\sqrt{3}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+70\sqrt{3}}{2*175}=\frac{0+70\sqrt{3}}{350}
=\frac{70\sqrt{3}}{350}
=\frac{\sqrt{3}}{5}
$