-10x2+490=0

Solution for -10x2+490=0 equation:

-10x^2+490=0

a = -10; b = 0; c = +490;
Δ = b2-4ac
Δ = 02-4·(-10)·490
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-140}{2*-10}=\frac{-140}{-20}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+140}{2*-10}=\frac{140}{-20}
=-7
$