430=0.1x^2+24x+25

Solution for 430=0.1x^2+24x+25 equation:

430=0.1x^2+24x+25

We move all terms to the left:

430-(0.1x^2+24x+25)=0

We get rid of parentheses

-0.1x^2-24x-25+430=0

We add all the numbers together, and all the variables

-0.1x^2-24x+405=0

a = -0.1; b = -24; c = +405;
Δ = b2-4ac
Δ = -242-4·(-0.1)·405
Δ = 738
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:

$\sqrt{\Delta}=\sqrt{738}=\sqrt{9*82}=\sqrt{9}*\sqrt{82}=3\sqrt{82}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-3\sqrt{82}}{2*-0.1}=\frac{24-3\sqrt{82}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+3\sqrt{82}}{2*-0.1}=\frac{24+3\sqrt{82}}{-0.2}
$