0,3x+0,012x^2=45

Solution for 0,3x+0,012x^2=45 equation:

0.3x+0.012x^2=45

We move all terms to the left:

0.3x+0.012x^2-(45)=0

a = 0.012; b = 0.3; c = -45;
Δ = b2-4ac
Δ = 0.32-4·0.012·(-45)
Δ = 2.25
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{-(0.3)-\sqrt{2.25}}{2*0.012}=\frac{-0.3-\sqrt{2.25}}{0.024}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{2.25}}{2*0.012}=\frac{-0.3+\sqrt{2.25}}{0.024}
$