-478(x)=4.9x^2+12

Solution for -478(x)=4.9x^2+12 equation:

-478(x)=4.9x^2+12

We move all terms to the left:

-478(x)-(4.9x^2+12)=0

We get rid of parentheses

-4.9x^2-478x-12=0

a = -4.9; b = -478; c = -12;
Δ = b2-4ac
Δ = -4782-4·(-4.9)·(-12)
Δ = 228248.8
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{-(-478)-\sqrt{228248.8}}{2*-4.9}=\frac{478-\sqrt{228248.8}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-478)+\sqrt{228248.8}}{2*-4.9}=\frac{478+\sqrt{228248.8}}{-9.8}
$