X^2+36x-98=0

Solution for X^2+36x-98=0 equation:

X^2+36X-98=0

a = 1; b = 36; c = -98;
Δ = b2-4ac
Δ = 362-4·1·(-98)
Δ = 1688
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:

$\sqrt{\Delta}=\sqrt{1688}=\sqrt{4*422}=\sqrt{4}*\sqrt{422}=2\sqrt{422}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{422}}{2*1}=\frac{-36-2\sqrt{422}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{422}}{2*1}=\frac{-36+2\sqrt{422}}{2}
$