32/10=2^2x

Solution for 32/10=2^2x equation:

32/10=2^2x

We move all terms to the left:

32/10-(2^2x)=0

We multiply all the terms by the denominator


-2^2x*10+32=0

Wy multiply elements

-20x^2+32=0

a = -20; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·(-20)·32
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*-20}=\frac{0-16\sqrt{10}}{-40}
=-\frac{16\sqrt{10}}{-40}
=-\frac{2\sqrt{10}}{-5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*-20}=\frac{0+16\sqrt{10}}{-40}
=\frac{16\sqrt{10}}{-40}
=\frac{2\sqrt{10}}{-5}
$