63=15+0.04x+0.003x^2

Solution for 63=15+0.04x+0.003x^2 equation:

63=15+0.04x+0.003x^2

We move all terms to the left:

63-(15+0.04x+0.003x^2)=0

We get rid of parentheses

-0.003x^2-0.04x-15+63=0

We add all the numbers together, and all the variables

-0.003x^2-0.04x+48=0

a = -0.003; b = -0.04; c = +48;
Δ = b2-4ac
Δ = -0.042-4·(-0.003)·48
Δ = 0.5776
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{-(-0.04)-\sqrt{0.5776}}{2*-0.003}=\frac{0.04-\sqrt{0.5776}}{-0.006}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.04)+\sqrt{0.5776}}{2*-0.003}=\frac{0.04+\sqrt{0.5776}}{-0.006}
$