16x2+8x-98=0

Solution for 16x2+8x-98=0 equation:

16x^2+8x-98=0

a = 16; b = 8; c = -98;
Δ = b2-4ac
Δ = 82-4·16·(-98)
Δ = 6336
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:

$\sqrt{\Delta}=\sqrt{6336}=\sqrt{576*11}=\sqrt{576}*\sqrt{11}=24\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-24\sqrt{11}}{2*16}=\frac{-8-24\sqrt{11}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+24\sqrt{11}}{2*16}=\frac{-8+24\sqrt{11}}{32}
$