0=10x^2+4x-98

Solution for 0=10x^2+4x-98 equation:

0=10x^2+4x-98

We move all terms to the left:

0-(10x^2+4x-98)=0

We add all the numbers together, and all the variables

-(10x^2+4x-98)=0

We get rid of parentheses

-10x^2-4x+98=0

a = -10; b = -4; c = +98;
Δ = b2-4ac
Δ = -42-4·(-10)·98
Δ = 3936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3936}=\sqrt{16*246}=\sqrt{16}*\sqrt{246}=4\sqrt{246}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{246}}{2*-10}=\frac{4-4\sqrt{246}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{246}}{2*-10}=\frac{4+4\sqrt{246}}{-20}
$