10x^2+4x-98=0

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

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}
$