100=0.25x^2+3x

Solution for 100=0.25x^2+3x equation:

100=0.25x^2+3x

We move all terms to the left:

100-(0.25x^2+3x)=0

We get rid of parentheses

-0.25x^2-3x+100=0

a = -0.25; b = -3; c = +100;
Δ = b2-4ac
Δ = -32-4·(-0.25)·100
Δ = 109
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{109}}{2*-0.25}=\frac{3-\sqrt{109}}{-0.5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{109}}{2*-0.25}=\frac{3+\sqrt{109}}{-0.5}
$