4x^2+38x-98=0

Solution for 4x^2+38x-98=0 equation:

4x^2+38x-98=0

a = 4; b = 38; c = -98;
Δ = b2-4ac
Δ = 382-4·4·(-98)
Δ = 3012
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:

$\sqrt{\Delta}=\sqrt{3012}=\sqrt{4*753}=\sqrt{4}*\sqrt{753}=2\sqrt{753}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{753}}{2*4}=\frac{-38-2\sqrt{753}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{753}}{2*4}=\frac{-38+2\sqrt{753}}{8}
$