50=7^x*5x+3

Solution for 50=7^x*5x+3 equation:

50=7^x*5x+3

We move all terms to the left:

50-(7^x*5x+3)=0

We get rid of parentheses

-7^x*5x-3+50=0

We add all the numbers together, and all the variables

-7^x*5x+47=0

Wy multiply elements

-35x^2+47=0

a = -35; b = 0; c = +47;
Δ = b2-4ac
Δ = 02-4·(-35)·47
Δ = 6580
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:

$\sqrt{\Delta}=\sqrt{6580}=\sqrt{4*1645}=\sqrt{4}*\sqrt{1645}=2\sqrt{1645}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1645}}{2*-35}=\frac{0-2\sqrt{1645}}{-70}
=-\frac{2\sqrt{1645}}{-70}
=-\frac{\sqrt{1645}}{-35}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1645}}{2*-35}=\frac{0+2\sqrt{1645}}{-70}
=\frac{2\sqrt{1645}}{-70}
=\frac{\sqrt{1645}}{-35}
$