(x)=-10x^2=490

Solution for (x)=-10x^2=490 equation:

(x)=-10x^2=490

We move all terms to the left:

(x)-(-10x^2)=0

We get rid of parentheses

10x^2+x=0

a = 10; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·10·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*10}=\frac{-2}{20}
=-1/10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*10}=\frac{0}{20}
=0
$