9x^2+(125^0.5x)+(25/4)=0

Solution for 9x^2+(125^0.5x)+(25/4)=0 equation:

9x^2+(125^0.5x)+(25/4)=0

We add all the numbers together, and all the variables

9x^2+(125^0.5x)+(+25/4)=0

We get rid of parentheses

9x^2+125^0.5x+25/4=0

We multiply all the terms by the denominator


9x^2*4+(125^0.5x)*4+25=0

We multiply parentheses

9x^2*4+500x+25=0

Wy multiply elements

36x^2+500x+25=0

a = 36; b = 500; c = +25;
Δ = b2-4ac
Δ = 5002-4·36·25
Δ = 246400
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:

$\sqrt{\Delta}=\sqrt{246400}=\sqrt{1600*154}=\sqrt{1600}*\sqrt{154}=40\sqrt{154}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(500)-40\sqrt{154}}{2*36}=\frac{-500-40\sqrt{154}}{72}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(500)+40\sqrt{154}}{2*36}=\frac{-500+40\sqrt{154}}{72}
$