Solve for x: 10^2x + 3(10^x) - 10 = 0

That would be great if you could show all the steps. Thanks! :D

10^(2x) = (10^x)²

Let t = 10^x so t > 0 because 10^x is always greater than 0

t² + 3t - 10 = 0
(t + 5)(t - 2) = 0
t = -5 is not the answer

t = 2

10^x = 2
take log

x = log2
x is about 0.301

10^2x can be rewritten

10²*10^x

let 10^x=V

the equation is now

10²V+3V-10=0

solve for V, replace V=10^x, solve for x

This is reducible to a quadratic.

Let u = 10^x

u^2 + 3u - 10 = 0
(u-2)(u+5) = 0
u = 2 or -5

10^x = 2 or 10^x = -5

Sorry, i don't know further from here since I havent done logarithms.