-0.1x^2+9x+21=-0.3x^2+18x+21

Solution for -0.1x^2+9x+21=-0.3x^2+18x+21 equation:

-0.1x^2+9x+21=-0.3x^2+18x+21

We move all terms to the left:

-0.1x^2+9x+21-(-0.3x^2+18x+21)=0

We get rid of parentheses

-0.1x^2+0.3x^2-18x+9x-21+21=0

We add all the numbers together, and all the variables

0.2x^2-9x=0

a = 0.2; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·0.2·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*0.2}=\frac{0}{0.4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*0.2}=\frac{18}{0.4}
=45
$