150=0.04x^2+1.1x

Solution for 150=0.04x^2+1.1x equation:

150=0.04x^2+1.1x

We move all terms to the left:

150-(0.04x^2+1.1x)=0

We get rid of parentheses

-0.04x^2-1.1x+150=0

a = -0.04; b = -1.1; c = +150;
Δ = b2-4ac
Δ = -1.12-4·(-0.04)·150
Δ = 25.21
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{-(-1.1)-\sqrt{25.21}}{2*-0.04}=\frac{1.1-\sqrt{25.21}}{-0.08}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.1)+\sqrt{25.21}}{2*-0.04}=\frac{1.1+\sqrt{25.21}}{-0.08}
$