25=(.003x^2)+(.62x)+3

Solution for 25=(.003x^2)+(.62x)+3 equation:

25=(.003x^2)+(.62x)+3

We move all terms to the left:

25-((.003x^2)+(.62x)+3)=0

We add all the numbers together, and all the variables

-((.003x^2)+(+.62x)+3)+25=0


We calculate terms in parentheses: -((.003x^2)+(+.62x)+3), so:

(.003x^2)+(+.62x)+3

We get rid of parentheses

.003x^2+.62x+3

Back to the equation:

-(.003x^2+.62x+3)


We get rid of parentheses

-.003x^2-.62x-3+25=0

We add all the numbers together, and all the variables

-0.003x^2-0.62x+22=0

a = -0.003; b = -0.62; c = +22;
Δ = b2-4ac
Δ = -0.622-4·(-0.003)·22
Δ = 0.6484
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{-(-0.62)-\sqrt{0.6484}}{2*-0.003}=\frac{0.62-\sqrt{0.6484}}{-0.006}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.62)+\sqrt{0.6484}}{2*-0.003}=\frac{0.62+\sqrt{0.6484}}{-0.006}
$