.006x^2+.3x=20

Solution for .006x^2+.3x=20 equation:

.006x^2+.3x=20

We move all terms to the left:

.006x^2+.3x-(20)=0

a = .006; b = .3; c = -20;
Δ = b2-4ac
Δ = .32-4·.006·(-20)
Δ = 0.57
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{0.57}}{2*.006}=\frac{-0.3-\sqrt{0.57}}{0.012}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.3)+\sqrt{0.57}}{2*.006}=\frac{-0.3+\sqrt{0.57}}{0.012}
$