16x^2-28x-98=0

Solution for 16x^2-28x-98=0 equation:

16x^2-28x-98=0

a = 16; b = -28; c = -98;
Δ = b2-4ac
Δ = -282-4·16·(-98)
Δ = 7056
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7056}=84$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-84}{2*16}=\frac{-56}{32}
=-1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+84}{2*16}=\frac{112}{32}
=3+1/2
$