100=0.3x^2+7x

Solution for 100=0.3x^2+7x equation:

100=0.3x^2+7x

We move all terms to the left:

100-(0.3x^2+7x)=0

We get rid of parentheses

-0.3x^2-7x+100=0

a = -0.3; b = -7; c = +100;
Δ = b2-4ac
Δ = -72-4·(-0.3)·100
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13}{2*-0.3}=\frac{-6}{-0.6}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*-0.3}=\frac{20}{-0.6}
=-33+0.2/0.6
$